SYNTHETIC 
INORGANIC  CHEMISTRY 


A  COURSE  OF  LABORATORY  AND  CLASSROOM 
STUDY  FOR  FIRST  YEAR  COLLEGE  STUDENTS 


BY 

ARTHUR  A.  BLANCHARD,  PH.D. 

Associate  Professor  of  Inorganic  Chemistry  at  the 
Massachusetts  Institute  of  Technology 

AND 

JOSEPH  W.  PHELAN,  S.B. 

Associate  Professor  of  Inorganic  Chemistry  at  the 
Massachusetts  Institute  of  Technology 


ENTIRELY  REWRITTEN  AND  GREATLY  ENLARGED 
THIRD   EDITION     TOTAL   ISSUE   EIGHT  THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:  CHAPMAN  &  HALL,  LIMITED 

1922 


COPYRIGHT,  1908,  1910,  1916, 

BY 

ARTHUR  A.  BLANCHARD 


COPYRIGHT,  1922, 

BY 
A.  A.  BLANCHARD 

AND 

J.  W.  PHELAN 


Stanbopc  press 

TECHNICAL  COMPOSITION  COMPANY 
F.  H.  GILSON  COMPANY 

BOSTON,  U.  S.  A.  10-22 


PREFACE  TO  THIRD  EDITION 

Since  the  adoption  of  an  entrance  requirement  in  chemistry 
at  the  Massachusetts  Institute  of  Technology  the  problem  of  the 
instruction  becomes  somewhat  simplified  in  that  a  certain  back- 
ground of  chemical  experience  is  possessed  by  all  the  students. 
Major  experiments  consisting  of  the  preparation  of  inorganic 
substances  are  still  believed  to  be  the  best  means  of  training  the 
chemical  students  in  the  laboratory,  and  the  present  edition  re- 
tains the  essential  features  of  the  former  edition  under  the  same 
title  prepared  by  one  of  the  authors.  It  has  seemed  very  desirable 
to  have  in  a  single  volume  a  comprehensive  plan  of  laboratory 
work  and  class  room  study  to  cover  a  whole  year's  work  and  to 
this  end  the  two  authors  have  put  together  the  material  which 
they  formerly  published  separately. 

Experience  has  taught  teachers  that  it  is  unwise  to  count  too 
much  upon  an  understanding  of  subjects  previously  studied. 
Therefore,  although  this  book  is  presumed  to  outline  a  course 
continuing  from  a  previous  course  in  high  school,  it  was  thought 
best  to  include  the  essential  material  of  the  high  school  course. 
This  elementary  material  was  presented  in  the  form  of  minor 
experiments  for  which  the  directions  were  written  as  if  the  ex- 
periments were  to  be  performed,  but  the  description  of  the  ob- 
servations and  the  discussion  of  the  chemical  principles  were  added. 
Thus  by  reading  the  text  the  pupil  could  quickly  obtain  a  grasp 
of  the  review  work.  This  treatment  of  the  minor  experiments 
however  proved  to  be  so  excellent  a  method  of  imparting  a  knowl- 
edge of  chemistry  that  it  was  followed  to  a  large  extent  through- 
out the  whole  book.  The  preparations  present  the  opportunity 
for  original  study;  the  minor  experiments  carry  along  the  system- 
atic instruction.  The  minor  experiments  can  too  at  all  times  be 
performed  when  it  seems  desirable. 

This  volume  is  therefore  considerably  more  than  a  laboratory 
manual;  the  authors  intend  to  make  it  the  text  book  of  the  full 
year's  work  in  inorganic  chemistry.  It  is  to  be  supplemented  by 
some  one  of  the  more  comprehensive  text  books,  which  is  to  be 


498929 


iy  PREFACE  TO  THIRD  EDITION 

used  for  reference  in  looking  up  the  purely  descriptive  parts  of 
chemistry  and  the  industrial  and  economic  aspects  of  the  subject. 
If  perhaps  the  authors  may  have  felt  at  times  a  certain  am- 
bition to  prepare  a  complete  text  book,  each  one  of  the  several 
excellent  books  of  this  scope  th,at  have  appeared  within  the  last 
few  years  has  shown  them  anew  the  entire  needlessness  of  under- 
taking such  a  task.  What  they  have  done,  namely  to  prepare 
a  systematic  course  of  study  of  the  facts  and  principles  of  chem- 
istry based  upon  the  work  of  the  laboratory,  is  rather  different 
from  anything  that  has  been  attempted  before. 

ARTHUR  A.  BLANCHARD. 
JOSEPH  W.  PHELAN. 
September,  1922 


PREFACE  TO  FIRST  EDITION 

THIS  series  of  notes  was  designed  to  serve  as  a  guide  for  labora- 
tory work  and  study  in  Inorganic  Chemistry  during  the  second 
term  of  the  first  year  at  the  Massachusetts  Institute  of  Technol- 
ogy. It  had  been  felt  for  some  time  that  Qualitative  Analysis, 
which  was  previously  made  the  basis  for  laboratory  practice  during 
that  period,  did  not  fully  meet  the  requirements  and  that  a  course 
based  upon  the  preparation  of  typical  chemical  substances  might 
prove  more  satisfactory.  In  consequence,  notes  in  essentially 
the  form  now  published  were  prepared  during  the  year  1906-07, 
they  being  the  direct  outcome  of  several  years'  previous  trial  of 
a  limited  amount  of  preparation  work.  The  present  book  is  a 
thorough  revision  of  those  notes  in  the  light  of  experience  in  their 
actual  application. 

During  the  first  term's  study  of  chemistry  there  can  be  little 
doubt  that  a  course  of  simple  experiments,  such  as  has  long  been 
in  use,  in  the  methods  of  formation  and  in  the  study  of  the  prop- 
erties of  the  non-metallic  elements  —  oxygen,  hydrogen,  the  halo- 
gens, sulphur,  nitrogen,  and  carbon  —  and  their  compounds,  is 
the  most  satisfactory.  But  when  it  comes  to  the  study  of  the 
metallic  elements,  three  options  as  to  laboratory  work  present 
themselves :  First,  a  continuation  of  experiments  similar  in  nature 
to  those  of  the  first  term;  second,  Qualitative  Analysis;  third, 
Preparation  Work.  The  disadvantages  of  the  first  plan  are  that 
the  experiments  are  so  quickly  performed  and  so  alike  in  charac- 
ter that  they  fail  to  arouse  much  enthusiasm  in  the  student  or  to 
leave  very  vivid  impressions  on  his  mind.  Qualitative  analysis 
is  in  many  ways  a  most  excellent  basis  for  teaching  the  chemistry 
of  the  metallic  elements;  but  its  chief  disadvantages  are:  First, 
that  it  is  one-sided,  it  dealing  as  it  does  almost  exclusively  with 
the  chemistry  of  solutions  and  the  formation  of  highly  insoluble 
substances;  second,  that  it  requires  the  sequence  followed  in  the 
lectures  to  be  that  of  the  qualitative  procedure  instead  of  a  more 
natural  one  based  on  the  periodic  classification;  and  third,  that 
it  is  well-nigh  impossible  to  keep  from  the  student's  mind  the  false 
idea  that  the  end  and  aim  of  qualitative  analysis  is  principally 
"to  get  the  unknowns  correct." 

v 


yi  PREFACE  TO  FIRST  EDITION 

Some  of  the  advantages  which  seem  to  be  possessed  by  a  course 
of  preparation  work  such  as  outlined  in  the  following  pages  are: 

1.  The  sequence  of  the  exercises  may  follow  that  of  the  lectures. 

2.  Very  varied  types  of  chemical  change  are  illustrated,  both 
those  in  the  furnace  a'nd  those  in  solution.     In  solution  advantage 
is  taken  not  only  of  high  degrees  of  insolubility,  but  also  of  differ- 
ences in  solubility  among  the  more  soluble  substances  as  well  as 
of  differences  in  the  effect  of  temperature  on  solubility. 

3.  The  danger  of  the  work  becoming  a  mechanical  following 
of  directions  is  reduced  by  the  introduction  of  study  questions  and 
experiments  with  each  exercise. 

4.  In  its  effect  in  awakening  the  student's  interest  this  line  of 
work  has  proved  particularly  successful,  —  the  making  of  prep- 
arations is,  in  fact,  in  its  very  nature  one  of  the  most  fascinating 
forms  of  chemical  work.     Since  each  preparation  requires  a  good 
deal  of  time  and  thought,  and  the  product  when  obtained  is 
something  definite  and  tangible,  the  knowledge  thus  gradually 
absorbed  is  more  definite  and  less  easily  forgotten  than  when 
the  laboratory  work  consists  of  a  large  number  of  test  tube  re- 
actions. 

After  the  completion  of  such  a  course  as  this,  if  the  student 
commences  analytical  work  with  some  conception  of  the  sources 
and  methods  of  obtaining  the  substances  which  he  is  to  use  as 
reagents,  etc.,  there  can  be  no  doubt  that  the  latter  work  will 
then  have  a  much  deeper  meaning. 

The  plan  kept  in  mind  in  preparing  this  course  is,  briefly,  as 
follows:  The  greater  part  of  the  preparations  selected  are  of 
industrial  importance,  and  for  the  starting  point  of  each  either 
natural  products  or  crude  manufactured  materials  are  used  so 
far  as  is  possible.  The  course  does  not  aim  to  be  an  exhaustive 
one  in  chemical  preparations,  but  a  limited  number  of  exercises 
are  selected  to  illustrate  the  most  important  types  of  compounds 
of  the  common  elements  and  the  most  important  methods.  Two 
or  three  times  as  many  exercises,  are  furnished  as  any  one  student 
will  be  able  to  complete  in  the  time  usually  allotted;  thus  differ- 
ent students  may  be  assigned  different  preparations. 

The   notes   for   each   exercise   are   divided   into   three   parts: 

I.  A  discussion  of  the  object  of  the  exercise,  with  an  outline  of 
the  principle  of  the  method  and  the  reasons  for  the  steps  involved. 

II.  Working  directions  which,  if  carefully  followed,  should  re- 


PREFACE  TO  FIRST  EDITION  vii 

suit  in  obtaining  a  satisfactory  product.  It  is  believed  far  better 
to  make  the  directions  very  explicit,  for  the  reason  that  the  in- 
experienced student  may  easily  become  discouraged  by  failures 
due  to  difficulties  which  he  is  unable  to  foresee.  Difficulties 
enough  are  sure  to  arise  to  develop  originality  and  resourcefulness. 
III.  Questions  for  study  which  involve  additional  laboratory 
experiments,  the  consulting  of  text  books,  and  original  reasoning. 

At  the  end  of  each  group  of  exercises  is  furnished  a  set  of  gen- 
eral study  questions,  and  this  arrangement  of  the  exercises  in 
groups  is  such  as  to  bring  out  the  relationships  shown  in  the 
periodic  classification  of  the  elements. 

In  the  discussions  and  questions  given  with  the  various  exer- 
cises it  is  assumed  that  the  student  has  an  elementary  knowl- 
edge of  the  electrolytic  dissociation  theory  and  of  the  principle 
of  mass  action.  In  the  opinion  of  the  author  a  great  opportunity 
is  lost  for  bringing  out  relationships  among  chemical  phenomena 
if  these  principles  are  not  taught  during  the  first  term's  study  of 
college  chemistry  and  their  applications  pointed  out  in  connection 
with  later  work  in  inorganic  preparations  and  in  analytical  chem- 
istry. The  effort  has  been  to  make  the  questions  such  as  cannot 
be  answered  mechanically.  Some  of  the  questions  may,  in  con- 
sequence, seem  rather  difficult  and  incapable  of  direct  answers; 
the  object  of  the  questions  is,  however,  not  solely  to  bring  forth 
correct  statements  of  facts  and  theories,  but  is  also  to  teach  the 
student  to  use  his  head  in  seeking  for  the  significance  of  facts  and 
in  reasoning  from  one  fact  to  another. 

Acknowledgment  is  due  to  many  sources  for  the  outline  of  the 
greater  part  of  the  methods  given.  The  details  of  all  of  them  have, 
however,  been  very  carefully  worked  over  and  adapted  for  the 
purpose  in  view. 

In  conclusion  the  author  wishes  to  express  his  sense  of  obli- 
gation to  Professor  Henry  P.  Talbot,  head  of  the  Department  of 
Chemistry,  at  whose  request  the  preparation  of  these  notes  was 
undertaken;  also  to  other  members  of  the  instructing  staff  at  the 
Massachusetts  Institute  of  Technology  for  helpful  criticism  and 
suggestions,  and  particularly  to  Professor  J.  W.  Phelan,  to  whose 
efficient  management  of  the  laboratory  instruction  is  due  any 
success  with  which  this  course  has  met  at  this  Institute. 

ARTHUR  A.  BLANCHARD. 

March,  1908. 


TABLE  OF  CONTENTS 


DIRECTIONS  FOR  WORK 1 

NOTES  ON  LABORATORY  MANIPULATION 4 

I.  Precipitation;    crystallization.     2.  Pouring.     3.  Transferring 
precipitates   or   crystals.     4.    Filtering;    collecting  precipitates. 
5.  Washing  precipitates.     6.  Evaporation.     7.  Dissolving  solid 
substances.     8.    Crystallization.     9.    Drying.     10.    Pulverizing. 

II.  Neutralizing.     12.  Dry  reactions;  furnaces.     13.  Gas  gener- 
ators. 

CHAPTER  I.   THE  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 
Experiments 22 

Weighing 23 

Exp.  1.   The  combining  Ratio  of  Zinc  and  Oxygen 24 

Exp.  2.   Weight  of  a  Liter  of  Oxygen 25 

Exp.  3.   Volume  of  Hydrogen  displaced  by  Zinc 26 

Exp.  4.   The  Combining  Ratio  of  Hydrogen  and  Oxygen 28 

Notes  and  Problems 30 

The  law  of  definite  proportions.  The  law  of  multiple  proportions. 
The  atomic  theory.  Atomic  weights.  Standard  of  atomic 
weights,  0  =  16.  Measurement  of  gases.  Boyle's  law.  Charles' 
law.  Dalton's  law.  Saturated  water  vapor.  Gay  Lussac's 
law  of  combining  volumes.  Avogadro's  principle.  Molecules  of 
elementary  gases.  Molecular  weights;  moles.  Molal  volume. 
Atomic  weights.  Derivation  of  a  formula. 

CHAPTER  II.   WATER  AND  SOLUTION 

Preparations 45 

1.  Potassium  Nitrate 45 

2.  Crystallized  Sodium  Carbonate,  Na^COs.  10  H2O 50 

3.  Potassium  and  Copper  Sulphate,  K2SO4.  CuSO4.  6  H20 52 

Experiments 53 

Hydrates.     Water   of   crystallization.     Composition    of   a   crystal 
hydrate.     Efflorescence.     Deliquescence 53 

Elements  and  Water.     Sodium.     Calcium.    Magnesium.    Iron.    Re- 
moval of  protective  coating  by  chemical  action.     Chlorine  and 

water 56 

iz 


X  TABLE  OF  CONTENTS 

Oxides  and  Water.     Sodium  oxide.     Calcium  oxide.     Magnesium 

oxide.     Non-metal  oxides 61 

Water  contains  two  separately  replaceable  portions  of  hydrogen 64 

Definitions 65 

Mole.  Molal  solution.  Equivalent.  Normal  solution.  Form- 
ula Weight.  Formal  solution."  Concentration. 

Problems 66 

General  Questions  II 67 

CHAPTER  III.  THE  THEORY  OP  IONIZATION 
Experiments 68 

Osmotic  pressure.  Electrical  conductivity  of  solutions.  Acids. 
Strong  and  weak  acids.  Bases.  Strong  and  weak  bases.  Neu- 
tralization of  a  strong  acid  and  a  strong  base.  Neutralization  of  a 
weak  acid  and  a  weak  base.  High  ionization  of  all  salt  solutions. 
Displacement  of  a  weak  acid.  Displacement  of  a  weak  base. 
Displacement  of  an  insoluble  base.  Characteristic  reactions  of 
certain  ions.  Electromotive  series.  Hydrolysis.  Solubility 
product.  Effect  of  its  neutral  salt  on  strength  of  a  weak  acid. 

Notes  and  Problems 79 

Measurement  of  Ionization.     Molal  lowering  of  the  freezing  point. 

Osmotic  pressure 79 

Ionization  Data 82 

Ionic  Reactions.  Ionization  a  reversible  reaction.  Equilibrium. 
Equations  for  Ionic  reactions.  Rules  for  writing  equations  in 

ionic  form 84 

Metathesis.  Precipitation.  Neutralization.  Neutralization  of  a 
weak  acid  and  a  weak  base.  Displacement  of  a  weak  acid.  Dis- 
placement of  a  weak  base.  Basic  properties  of  metal  hydroxides. 

Formation  of  volatile  products 87 

Complexions.     Ammoniates.     Complex  negative  ions 97 

Hydrolysis 101 

Ionization  of  Polybasic  Acids 102 

Reactions  of  Oxidation  and  Reduction.     Electromotive  series 104 

Law  of  Molecular  Concentration 106 

Solubility  and  Solubility  Product Ill 

CHAPTER  IV.   THE  NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

Preparations • 114 

4.  Copper  Oxide 114 

5.  Hydrogen  Peroxide  and  Barium  Peroxide  Hydrate 115 

6.  Hydrochloric. Acid 117 

7.  Hydrobromic  Acid 120 


TABLE  OF  CONTENTS  xi 

8.  Hydriodic  Acid 122 

9.  Barium  Bromide 124 

10.  Aluminum  Sulphide 125 

11.  Calcium  Sulphide 127 

12.  Mercuric  Sulphide 128 

13.  Aluminum  Nitride 129 

14.  Magnesium  Nitride  and  Ammonia 131 

Experiments 134 

Valence.  Oxides.  Peroxides.  The  halogens.  Formation  and 
properties  of  the  hydrogen  halides.  Characteristic  reaction  of 
the  halide  ions.  Relative  activity  of  the  halogens,  oxygen  and 
sulphur.  Sulphur.  Nitrogen. 

General  Questions  IV 157 


CHAPTER  V.  ALKALI  AND  ALKALINE  EARTH  METALS 

Preparations 158 

15.  Sodium  Bicarbonate  by  the  Ammonia  Process 158 

16.  Sodium  Carbonate  from  Sodium  Bicarbonate. 160 

17.  Caustic  Alkali  from  Alkali  Carbonate 160 

18.  Chemically  Pure  Sodium  Chloride  from  Rock  Salt 163 

19.  Ammonium  Bromide 165 

20.  Strontium  Hydroxide  from  Strontium  Sulphate 167 

21.  Strontium  Chloride  from  Strontium  Sulphate 169 

22.  Barium  Oxide  and  Barium  Hydroxide 172 

Experiments .  174 

Stability  of  carbonates.  Oxides  and  water.  Solubility  and 
basic  strength  of  hydroxides.  Ammonium  compounds. 

General  Questions  V 176 

CHAPTER  VI.   ELEMENTS  OF  GROUP  III 

Preparations 179 

23.  Boric  Acid 179 

24.  Alum  from  Cryolite 181 

25.  Alum  and  Ammonia  from  Aluminum  Nitride 185 

26.  Hydrated  Aluminum  Chloride,  AlCla.  6  H20 187 

27.  Anhydrous  Aluminum  Bromide 189 

Experiments 191 

Acid  strength  of  boric  acid.  Amphoteric  substances.  Acid  and 
basic  strength  of  aluminum  hydroxide.  Hydrolysis  of  aluminum 
carbonate. 

General  Questions  VI 194 


TABLE  OF  CONTENTS 

CHAPTER  VII.   HEAVY  METALS  OF  GROUPS  I  AND  II 

Preparations 196 

28.  Crystallized  Copper  Sulphate  from  Copper  Turnings 196 

29.  Cuprous  Chloride 198 

30.  Cuprous  Oxide ., 200 

31.  Ammonio-Copper  Sulphate,  CuSO4.  4  NH3.  H2O 202 

32.  Zinc  Oxide 204 

33.  Mercurous  Nitrate 206 

34.  Mercuric  Nitrate 207 

35.  Mercuric  Sulphocyanate 208 

Experiments 210 

Stability  of  carbonates.  Hydrolysis  of  salts.  Hydroxides. 
Basic  strength  of  silver  oxide.  Ammoniates.  Complex  negative 
ions.  Sulphides.  Electromotive  series. 

General  Questions  VII 214 

CHAPTER  VIII.   THE  OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

Preparations 216 

36.  Potassium  Bromate  and  Potassium  Bromide 217 

37.  Potassium  Chlorate 220 

38.  Potassium  lodate 222 

39.  lodic  Acid;  Iodine  Pentoxide 223 

40.  Potassium  Perchlorate 225 

41.  Sodium  Thiosulphate 226 

Experiments 229 

Hypochlorites.  Hypobromites.  Chlorates  and  bromates.  Bromic 
and  iodic  acids.  Properties  of  potassium  chlorate.  Reduction  of 
iodicacid.  Sulphur  dioxide.  Sulphurous  acid.  Reducing  action 
of  sulphurous  acid.  Oxidizing  action  of  sulphur  dioxide  and  sul- 
phurous acid.  Dehydrating  action  of  sulphuric  acid.  Oxidizing 
action  of  sulphuric  acid.  Nitric  acid  as  an  oxidizing  agent. 
Nitrous  acid.  Reducing  action  of  nitrous  acid. 

General  Questions  VIII 237 

CHAPTER  IX.   ELEMENTS  OF  GROUP  IV 

Preparations 239 

42.  Precipitated  Silica 239 

43.  Stannous  Chloride,  SnCl2.  2  H20 241 

44.  Stannic  Sulphide  (Mosaic  Gold) 244 

45.  Stannic  Chloride  (Anhydrous) 246 

46.  Lead  Nitrate 248 

47.  Lead  Dioxide 249 

48.  Red  Lead. .  251 


TABLE  OF  CONTENTS  xiii 

Experiments 252 

Carbon  dioxide.  Combustibility  of  carbon  compounds.  Carbon 
monoxide.  Carbides.  Luminosity  of  flame.  Silicon  dioxide 
and  silicic  acid.  Hydrolysis  of  stannous  salts.  Lead  salts. 
Amphoteric  character  of  hydroxides  of  tin  and  lead.  Stannic 
acid.  Thio-salts  of  tin.  Lead  dioxide.  Lead  tetrachloride. 
Stability  of  lead  carbonate. 

General  Questions  IX 264 

CHAPTER  X.   ELEMENTS  OF  GROUP  V 

Preparations 266 

49.  Ortho-Phosphoric  Acid 266 

50.  Crystallized  Arsenic  Acid 269 

51.  Antimony  Trichloride 272 

52.  Sodium  Sulphantimonate 274 

53.  Antimony  Pentasulphide 276 

54.  Metallic  Antimony 277 

55.  Bismuth  Basic  Nitrate 278 

Experiments : 279 

Oxidation  products  of  the  metals.  Sulphides  and  thio-salts. 
Reducing  action  of  phosphorous  acid.  Non-oxidizing  property  of 
phosphoric  acid.  Arsenious  and  arsenic  acids.  Reduction  of 
bismuth  salts.  Bismuth  in  a  higher  state  of  oxidation. 

General  Questions  X 282 

CHAPTER  XI.   HEAVY  METALS  OF  GROUPS  VI,  VII,  AND  VIII 

Preparations 285 

56.  Potassium  Bichromate 285 

57.  Potassium  Chromate 287 

58.  Chromic  Anhydride. 288 

59.  Ammonium  Bichromate 289 

60.  Chromic  Alum 290 

61.  Chromium  Metal 292 

62.  Manganous  Chloride 294 

63.  Potassium  Permanganate 296 

64.  Manganese  Metal 298 

65.  Ferrous  Ammonium  Sulphate 299 

66.  Ferric  Alum 300 

Experiments 302 

Stability  of  carbonates  of  metals  in  divalent  state.  To  show 
whether  the  carbonate  of  a  trivalent  metal  can  exist.  Oxidation 
of  a  divalent  oxide.  Properties  of  the  hydroxides.  Action  of 


xiv  TABLE  OF  CONTENTS 

alkaline  oxidizing  agents  on  the  hydroxides.     Oxidation  in  alka- 
line fusion.     Permanganate.     Chromate  and  dichromate. 

General  Questions  XI 307 

APPENDIX 

Concentration  of  Reagents ? 309 

Tension  of  Saturated  Aqueous  Vapor 310 

Electromotive  Series 310 

Chart  Atomic  Weights  and  Atomic  Numbers 311 

PERIODIC  CLASSIFICATION  OF  THE  ELEMENTS  ACCORDING  TO  THEIR 
ATOMIC  NUMBERS  AND  THE  ARRANGEMENT  OF  THEIR  ELECTRONS: 

Atomic  Number;  Theory  of  the  Structure  of  the  Atom;   Chart.  .  312 

Solubility  Tables 315 

TABLE  OF  ATOMIC  WEIGHTS Inside  front  cover 

PERIODIC  ARRANGEMENT  OF  THE  ELEMENTS  . .  Inside  back  cover 


SYNTHETIC 
INORGANIC  CHEMISTRY 


DIRECTIONS  FOR  WORK 

The  course  outlined  in  this  book  is  an  experimental  study  of 
chemistry.  Chapters  I  and  III  deal  with  general  principles. 
The  first  part  of  each  of  these  two  chapters  gives  directions  for 
experiments  which  are  to  be  performed  by  the  student.  Records 
of  these  experiments  are  to  be  kept  in  the  laboratory  note  book  as 
follows:  the  experimental  facts  and  measurements  are  to  be  writ- 
ten on  the  left  hand  page  as  the  note  book  lies  open;  opposite 
these  statements,  on  the  right  hand  page,  calculations  are  to  be 
made,  equations  for  the  chemical  reactions  are  to  be  written, 
and  final  conclusions  are  to  be  drawn.  The  second  part  of  each 
of  these  chapters  is  devoted  to  notes  discussing  the  principles 
that  the  experiments  illustrate,  and  problems  for  home  work. 

The  other  nine  chapters  are  devoted  to  preparations  and  ex- 
periments which  reveal  the  properties  of  the  various  classes  of  the 
chemical  elements. 

Preliminary  Reports  on  the  Preparations.  —  Before  beginning 
work  on  a  preparation  the  student  should  have  a  clear  knowledge 
of  the  whole  procedure  and  should  understand  the  reactions  as  well 
as  the  application  of  chemical  principles  to  these  reactions. 

To  that  end  study  carefully  the  general  discussion  of  the  prep- 
aration as  well  as  the  procedure.  Then  write  in  the  note  book  all 
reactions,  and,  starting  with  the  given  amount  of  the  principal 
raw  material,  calculate  what  amounts  of  the  other  substances 
are  necessary  to  satisfy  the  equations.  When  the  amount  speci- 
fied in  the  directions  is  different  from  that  calculated,  state  the 
reason  for  the  difference.  Calculate  also  on  the  basis  of  the 
equations  the  amount  of  the  main  product  as  well  as  of  any  im- 
portant intermediate  products  or  by-products. 

Present  this  preliminary  report  to  an  instructor  and  obtain 
his  approval  before  beginning  operations. 

1 


2  DIRECTIONS  FOR  WORK 

Manipulation.  —  All  references  from  the  procedure  to  the 
general  notes  on  laboratory  manipulation  (pp.  4-21)  should  have 
been  studied  before  making  the  preliminary  report.  Indeed  the 
instructor  will  probably  make  sure  by  a  quiz  that  this  has  been 
done  before  he  accepts  the  preliminary  report. 

Laboratory  Record.  —  The  working  directions,  in  the  section 
entitled  procedure,  are  to  be  kept  at  hand  while  carrying  out  the 
manipulations.  These  directions  do  not  need  to  be  copied  in  the 
laboratory  note  book;  but  it  is  essential,  nevertheless,  to  keep  a 
laboratory  record  in  which  are  entered  all  important  observations 
and  data;  such  as,  for  example,  appearance  of  solutions  (color, 
turbidity);  appearance  of  precipitates  or  crystals  (color,  size  of 
grains,  crystalline  form);  results  of  all  weighings  or  measure- 
ments; number  of  recrystallizations;  results  of  test  for  purity 
of  materials  and  products,  etc. 

Questions.  —  The  section  under  this  title  gives  suggestions 
for  study,  which  involves  laboratory  experiments,  consultation 
of  reference  books,  of  which  all  that  are  necessary  will  be  found 
upon  the  shelf  in  the  laboratory,  and  reasoning. 

The  answers  to  the  questions  should  be  written  in  the  labora- 
tory note  book  following  the  entries  for  the  exercise,  and  this 
book  should  be  submitted  at  the  same  time  as  the  preparation  for 
the  approval  of  an  instructor. 

Use  of  Time  in  Laboratory.  —  In  preparation  work  it  is  fre- 
quently necessary  to  wait  for  considerable  periods  of  tune  for 
evaporations,  crystallizations,  etc.,  to  take  place.  This  time  may 
be  utilized  for  work  upon  the  study  questions  and  experiments, 
but  even  then  it  is  advisable  to  have  usually  more  than  a  single 
preparation  under  way.  Thus  no  time  need  be  wasted  by  the 
energetic  student  who  plans  his  work  well. 

Yield  of  Product.  —  Where  possible  the  methods  employed  in 
these  preparations  resemble  those  actually  used  on  an  industrial 
scale;  where  this  is,  however,  impossible  on  the  limited  scale  of 
the  laboratory,  mention  is  made  of  the  fact,  with  reasons  therefor. 
On  account  of  the  limitations  connected  with  work  on  a  labora- 
tory scale,  it  is  of  course  impossible  to  get  as  high  percentage  yields 
as  could  be  obtained  on  a  commercial  scale.  The  amounts  ob- 
tained of  each  preparation  are  to  be  weighed  and  recorded,  but 
the  chief  stress  is  to  be  laid  upon  the  excellence  of  the  product 
rather  than  upon  its  quantity. 


DIRECTIONS  FOR  WORK  3 

Experiments.  —  The  second  part  of  each  of  the  nine  chapters, 
of  which  the  preparations  comprise  the  first  part,  is  devoted  to 
short  experiments.  Not  only  are  the  directions  for  these  experi- 
ments given,  but  the  results  to  be  observed  are  stated,  and  the 
meaning  of  the  results  is  discussed.  Thus  this  experimental  part 
may  be  studied,  and  the  experiments  may  or  may  not  be  actually 
performed  according  to  the  discretion  of  the  student,  or  the  ad- 
vice of  the  instructor.  The  study  of  this  part  should  be  made  by 
every  student  as  a  preparation  for  the  Report  which  he  is  expected 
to  write  on  the  chemistry  of  the  elements  dealt  with  in  the  chap- 
ter. 

General  Questions.  —  These  questions  which  appear  at  the  end 
of  each  of  the  nine  chapters  are  to  serve  as  the  basis  of  the  written 
report  referred  to  in  the  preceding  paragraph. 

Number  of  Preparations.  —  A  certain  number  of  the  prep- 
arations will  be  designated  each  term  as  " required"  which  means 
that  they  will  be  discussed  in  detail  in  the  class  room  and  that 
detailed  knowledge  of  them  will  be  assumed  when  examination 
questions  are  made  out.  Besides  the  required  preparations, 
students  will  be  able  to  make  a  number  of  other  ones  of  their  own 
selection,  —  this  selection  of  -course  being  subject  to  the  instruc- 
tor's approval,  since  the  laboratory  may  not  at  all  times  have  the 
necessary  materials  and  apparatus. 


NOTES  ON  LABORATORY  MANIPULATION 

THESE  notes  are  intended  to  help  the  student  in  foreseeing 
and  in  overcoming  some  of  the  difficulties  that  arise  in  experi- 
mental work.  They  by  no  means  make  it  unnecessary  for  him 
to  exercise  ingenuity  and  originality  in  planning  and  carrying  out 
the  details  of  laboratory  work.  At  the  outset  these  notes  should 
be  read  through  carefully;  then  when  in  the  later  work  references 
to  specific  notes  are  made  their  general  bearing  will  be  better 
appreciated. 

1.    PRECIPITATION;  CRYSTALLIZATION 

In  the  majority  of  chemical  processes  which  are  carried  out 
in  the  wet  way,  separations  are  accomplished  by  taking  advantage 
of  differences  in  solubility.  In  case  a  certain  product  is  extremely 
insoluble  and  is  formed  almost  instantaneously  when  solutions 
containing  the  requisite  components  are  mixed,  the  process  is  called 
precipitation  and  the  insoluble  substance  is  called  the  precipitate. 
If  the  product  to  be  formed  is  less  insoluble,  so  that  it  separates 
more  slowly,  or  only  after  evaporating  away  a  part  of  the  solvent, 
the  process  is  called  crystallization. 

In  some  cases  the  precipitate,  or  the  crystals,  constitute  the 
desired  product;  in  other  cases  a  product  which  it  is  necessary  to 
remove  from  the  solution  before  the  desired  product  can  be  ob- 
tained pure.  In  either  case  it  is  necessary  to  make  as  complete 
a  separation  as  possible  of  the  solid  from  the  liquid.  This  in- 
volves the  manipulations  described  under  Notes  2,  3,  and  4. 

2.    POURING 

In  pouring  a  liquid  from  a  vessel,  either  into  a  filter  or  into 
another  vessel,  care  must  be  taken  not  to  slop  the  liquid  nor  to 
allow  it  to  run  down  the  outside  of  the  vessel  from  which  it  is 
poured.  To  this  end  touch  a  stirring  rod  to  the  lip  of  the  dish 
or  beaker  (Fig.  1)  and  allow  the  liquid  to  run  down  the  rod. 

4 


FILTERING;    COLLECTING  PRECIPITATES  5 

3.    TRANSFERRING  PRECIPITATES  OR  CRYSTALS 

If  large  crystals  have  separated  from  a  liquid  they  may  be  picked 
out,  or  the  liquid  may  be  poured  off. 

If  a  precipitate  or  a  crystalline  meal  has  formed  it  must  be 
drained  in  a  filter  funnel.  First  pour  off  the  liquid  (see  Note  2)  — 
through  the  filter  if  necessary,  so  as  to  save  any  floating  particles 
of  the  solid  —  then  pour  the  main  part  of  the  damp  solid  into  the 


FIG.  1  FIG.  2 

filter.  A  considerable  part  of  the  solid  will  adhere  to  the  dish; 
most  of  this  may  be  scraped  out  by  means  of  a  spatula,  but  the 
last  of  it  is  most  easily  rinsed  into  the  filter.  For  rinsing,  a  jet  of 
water  from  the  wash  bottle  (Fig.  2)  may  be  used  if  the  solid  is  very 
insoluble.  If  the  solid  is  soluble  in  water,  some  of  the  saturated 
solution  may  be  poured  back  into  the  dish  from  out  of  the  filter 
bottle,  and  by  means  of  this  the  last  of  the  solid  may  be  removed 
to  the  filter. 

4.    FILTERING;   COLLECTING  PRECIPITATES 

(a)  A  coarse-grained  crystal  meal  can  best  be  collected  in  a  filter 
funnel  in  which  a  perforated  porcelain  plate  is  placed,  and  the 
mother  liquor  clinging  to  the  crystals  can  best  be  removed  with 
the  aid  of  suction  (see  next  paragraph). 


6 


NOTES  ON  LABORATORY  MANIPULATION 


(6)  Filtering  with  Suction.  —  With  a  fine-grained  crystal  meal, 
or  a  precipitate  which  is  not  of  such  a  slimy  character  as  to  clog 
the  pores  of  the  filter  paper,  a  suction  filter  is  most  advantageously 
used.  A  5-inch  filter  funnel  should  be  fitted  tightly  by  means  of 
a  rubber  stopper  into  the  neck  of  a  500  cc.  filter  bottle  (Fig.  3). 


FIG.  3 

Place  a  Ij-inch  perforated  filter  plate  in  the  funnel  and  on  this  a 
disk  of  filter  paper  cut  so  that  its  edges  will  turn  up  about  3  mm. 
on  the  side  of  the  funnel  all  the  way  around.  Hold  the  disk  of 
dry  paper  in  the  right  position,  wet  it  with  a  jet  from  the  wash 
bottle,  draw  it  firmly  down  against  the  filter  plate  by  applying 
the  suction,  and  press  the  edges  firmly  against  the  side  of  the 
funnel,  so  that  no  free  channel  shall  remain.  In  pouring  the 
liquid,  direct  it  with  a  stirring  rod  (Note  2)  onto  the  middle  of  the 
filter;  do  not  allow  it  to  run  down  the  side  of  the  funnel,  as  this 
might  turn  up  the  edge  of  the  paper  and  allow  some  of  the  pre- 
cipitate to  pass  by.  After  bringing  all  of  the  solid  upon  the  filter 
it  may  be  freed  from  a  large  part  of  the  adhering  liquid  by  means 
of  the  suction,  and  it  may  then  be  purified  by  washing  with  a 
suitable  liquid  (see  Note  5) . 

The  suction  filter  is  very  generally  useful  for  the  purpose  of 
separating  a  solid  product  from  a  liquid.  In  cases  that  the  liquid 
runs  slowly,  the  rate  of  filtration  can  be  increased  by  using  a  larger 


FILTERING;    COLLECTING  PRECIPITATES  7 

filter  plate  or  still  better  a  Biichner  funnel  and  thereby  increasing 
the  filtering  area.  The  student  should,  however,  avoid  using  the 
suction  indiscriminately,  for  in  many  cases  it  is,  as  explained  in 
paragraph  c,  a  positive  disadvantage. 

Suction.  —  The  most  convenient  source  of  suction  is  the  Rich- 
ards water  pump,  which  can  be  attached  directly  to  the  water  tap. 
If  the  water  is  supplied  at  a  pressure  of  somewhat  over  one  at- 
mosphere (34  feet  of  water),  a  vacuum  of  very  nearly  an  atmos- 
phere can  be  obtained.  If  the  pressure  is  insufficient,  an  equally 
good  vacuum  can  be  obtained  by  means  of  the  suction  of  the  escap- 
ing water.  To  this  end  the  escape  pipe  must  be  prolonged  by  a 
tube  sufficiently  constricted  to  prevent  the  sections  of  the  descend- 
ing water  column  from  breaking  and  thus  allowing  air  to  enter 
from  the  bottom. 

To  keep  the  suction  pump  working  continuously,  however,  is 
extravagant  of  water  as  well  as  being  a  nuisance  in  the  laboratory 
on  account  of  the  unnecessary  noise.  Consequently  this  rule  is 
made  and  must  be  observed: 

The  suction  pump  must  never  be  kept  in  operation  more  than  two 
minutes  at  one  time. 

If  suction  must  be  applied  for  more  than  that  length  of  time, 
the  vacuum  which  is  produced  inside  of  the  two  minutes  may  be 
maintained  in  the  suction  bottle  by  closing  the  screw  cock.  (See 
Fig.  3.)  Thus,  if  all  the  joints  of  the  bottle  are  tight,  a  slimy 
precipitate  may  be  left  filtering  under  suction  over  night,  or  even 
longer. 

Trap.  —  The  use  of  the  trap  shown  in  the  diagram  is  most 
necessary,  as  otherwise  dirty  water  may  be  sucked  back  acciden- 
tally and  contaminate  the  solution  in  the  filter  bottle. 

(c)  Filtering  without  Suction.  —  A  slimy  or  gelatinous  precipitate 
can  be  collected  much  better  without  suction.  Suction  drags  the 
solid  matter  so  completely  into  the  pores  of  the  filter  that  in  most 
cases  the  liquid  nearly  ceases  to  run.  A  filter  funnel  and  filter 
should  be  chosen  large  enough  to  hold  the  entire  precipitate.  The 
filter  paper  should  be  folded  twice  and  then  opened  out  in  the  form 
of  a  cone  and  fitted  into  the  funnel.  The  corners  of  the  filter 
should  be  cut  off  round,  and  the  upper  edge  of  the  filter  should 
come  about  one-half  inch  below  the  rim  of  the  funnel.  It  is  best 
to  fit  the  paper  carefully  into  the  funnel,  to  wet  it  and  press  it  up 
against  the  glass  all  around,  so  that  there  will  be  no  air  channels. 


8        NOTES  ON  LABORATORY  MANIPULATION 

In  the  case  of  slow-running  liquids,  if  a  large  filter  is  used,  it 
may  be  filled  at  intervals  and  left  to  take  care  of  itself  the  rest  of 
the  time  while  other  work  is  being  done. 

In  case  a  considerable  weight  of  liquid  is  to  come  on  the  point 
of  the  filter,  this  may  be  reertforced  by  means  of  a  piece  of  linen 
cloth,  which  should  be  placed  under  the  middle  of  the  filter  paper 
before  it  is  folded,  and  should  then  be  folded  in  with  it  so  as  to 
strengthen. the  point. 

After  the  precipitate  is  collected  in  the  filter  and  drained,  it 
should  if  necessary  be  washed  (  see  Note  5). 

Both  filtration  and  washing  take  place  much  more  rapidly  if  the 
liquid  is  hot.  Time  can  also  usually  be  saved  if  the  precipitate  is 
allowed  to  settle  as  completely  as  possible  before  commencing  to 
filter.  The  clear  liquid  can  then  be  decanted  off,  or  if  necessary 
poured  rapidly  through  the  filter  before  the  latter  becomes  clogged 
with  the  main  part  of  the  precipitate. 

(d)  Filtering  Corrosive  Liquids.  —  Solutions  of  very  strong 
oxidizing  agents,  concentrated  solutions  of  the  strong  acids  and 
bases,  and  concentrated  solutions  of  a  few  salts  of  the  heavy 
metals  —  notably  zinc  chloride  and  stannous  chloride  —  attack 
filter  paper.  Ordinary  paper  is  thus  unserviceable  for  filtration, 
but  a  felt  made  of  asbestos  fibers  is  in  many  cases  very  useful. 
Shredded  asbestos,  which  has  been  purified  by  boiling  with  hydro- 
chloric acid  and  subsequent  washing,  is  suspended  in  water; 
the  suspension  is  poured  onto  a  perforated  plate  placed  in  a  filter 
funnel ;  and  suction  is  applied  whereby  the  water  is  removed  and 
the  fibers  are  drawn  together  to  form  a  compact  felt  over  the  filter 
plate.  Enough  asbestos  should  be  used  to  make  a  felt  1  to  3  mm. 
thick,  and  care  must  be  taken  to  see  that  it  is  of  uniform  thickness 
and  that  no  free  channels  are  left  through  which  solid  matter 
may  be  drawn.  Before  it  is  ready  for  use  a  considerable  amount 
of  water  should  be  drawn  through  the  filter,  and  the  loose  fibers 
should  be  rinsed  out  of  the  filter  bottle.  Before  pouring  the  liquid 
onto  the  filter  the  suction  should  be  started  gently,  and  the  liquid 
should  be  directed  by  means  of  a  stirring  rod  (Note  2)  onto  the 
middle  of  the  filter.  If  these  precautions  are  not  observed  the 
felt  may  become  turned  up  in  places,  so  that  the  precipitate  will 
pass  through. 

A  wad  of  glass  wool  in  the  bottom  of  a  glass  funnel  may  some- 
times be  used  to  strain  corrosive  liquids. 


WASHING  PRECIPITATES  9 

(e)  Cloudy  Filtrates.  —  When  a  filtrate  at  first  comes  through 
cloudy,  it  is  usually  sufficient  to  pour  the  first  portion  through  the 
filter  a  second  time.  The  pores  of  the  filter  soon  become  partially 
closed  with  the  precipitate,  so  that  then  even  the  finest  particles 
are  retained.  With  some  very  fine-grained  precipitates,  repeat- 
edly pouring  the  filtrate  through  the  same  filter  will  finally  give  a 
clear  filtrate. 

Special  kinds  of  filter  paper  are  made  to  retain  very  fine  precipi- 
tates, but  they  allow  the  liquid  to  pass  much  more  slowly  than 
ordinary  filters,  and  their  use  is  by  no  means  essential  in  any  of  the 
following  preparations. 

(/)  To  Keep  Liquids  Hot  during  Filtration.  —  When  liquids 
must  be  kept  hot  during  a  slow  filtration,  as,  for  example,  when 
cooling  would  cause  a  separation  of  crystals  that  would  clog  the 
filter,  it  sometimes  becomes  necessary  to  surround  the  funnel  with 
a  jacket  which  is  heated  with  steam  or  boiling  water.  In  the 
following  preparations  the  use  of  such  a  device  will  not  be  neces- 
sary, although  there  are  several  instances  where  it  is  necessary  to 
work  quickly  to  avoid  clogging  the  filter. 

(g)  Cloth  Filters.  —  In  preparations  made  on  a  small  scale,  paper 
filters  placed  in  ordinary  filter  funnels  are  invariably  used  if  the 
liquid  is  not  too  corrosive.  On  a  larger  scale  or  in  commercial 
practice,  cloth  is  much  used  for  filters,  and  it  can  be  made  in  the 
shape  of  bags  or  it  can  be  stretched  over  wooden  frames.  The 
cloth  or  other  filtering  medium  (asbestos,  paper  pulp,  sand,  etc.) 
has  to  be  chosen  in  each  case  with  reference  to  the  nature  of  the 
precipitate  and  the  corrosiveness  of  the  liquid. 

Many  of  the  preparations  in  this  book,  if  carried  out  on  a  larger 
scale  than  given  in  the  directions,  would  require  the  use  of  such 
cloth  filters.  It  is  often  advantageous  to  tack  one  piece  of  cloth 
permanently  across  a  wooden  support  and  on  top  of  this  to  lay  a 
second  cloth.  The  precipitate  can  then  be  easily  removed  together 
with  the  unfastened  cloth. 

For  devices  for  rapid  filtration  and  filtration  in  general  on  a  large 
scale,  a  work  on  Industrial  Chemistry  should  be  consulted. 

5.    WASHING  PRECIPITATES 

(a)  Washing  on  the  Filter.  —  To  remove  completely  the  im- 
purities contained  in  the  mother  liquor  clinging  to  precipitates  or 
crystals,  the  solid  is  washed.  Pure  water  is  used  for  washing, 


10 


NOTES  ON  LABORATORY  MANIPULATION 


provided  the  solid  is  not  too  soluble  or  is  not  decomposed  (hydro- 
lyzed)  by  it.  Special  directions  will  be  given  when  it  is  necessary 
to  use  other  than  pure  water. 

First,  the  solid  should  be  allowed  to  drain  as  completely  as 
possible,  then  the  wash  liquid  should  be  applied,  preferably  from 
the  jet  of  a  wash  bottle,  so  as  to  wet  the  whole  mass  and  to  rinse 
down  the  sides  of  the  filter.  If  suction  is  used,  suck  the  solid  as 
dry  as  possible,  then  stop  the  suction  while  applying  the  washing 
liquid;  after  the  solid  is  thoroughly  wet,  suck  out  the  liquid  and 
repeat  the  washing. 

A  little  thought  will  make  it  clear  that  the  washing  is  much 
more  effective  if  the  liquid  is  removed  as  completely  as  possible 
each  time  before  applying  fresh  wash  liquid,  and  that  a  number  of 
washings  with  a  small  amount  of  liquid  each  time  is  more  effective 
than  fewer  washings  with  much  greater  quantities  of  wash  liquid. 
It  is,  of  course,  evident  that  with  each  washing  the  liquid  should 
penetrate  to  all  parts  of  the  solid  material. 

(6)  Washing  by  Decantation.  —  In  case  a  precipitate  is  very 
insoluble  it  can  be  most  thoroughly  and  quickly  washed  by  de- 
cantation.  This  consists  in  allowing  it  to 
settle  in  a  deep  vessel  and  then  in  pouring 
(decanting)  or  siphoning  off  the  clear  liquid. 
Following  this  the  precipitate  is  stirred  up 
with  fresh  water  and  allowed  to  settle,  and 
the  liquid  is  again  decanted  off.  By  a  suffi- 
cient number  of  repetitions  of  this  process, 
the  precipitate  may  be  washed  entirely  free 
from  any  soluble  impurity,  after  which  ifc 
may  be  thrown  on  a  filter,  drained,  and  then 
dried. 

Most  precipitates,  even  after  they  have 
settled  as  completely  as  possible  in  the  liquid 
from  which  they  were  thrown  down,  are  very 
bulky,  and  their  apparent  volume  is  very 
large  as  compared  with  the  actual  volume  of 
the  solid  matter  itself.  For  example,  a 
precipitate  of  basic  zinc  carbonate  (Prep. 
32),  after  it  has  settled  as  completely  as  possible  in  a  deep 
jar  (Fig.  4),  may  still  occupy  a  volume  of  400  cc.  When  this 


FIG.  4 


EVAPORATION  11 

bulky  precipitate  is  dried,  however,  it  shrivels  up  into  a  few 
small  lumps  whose  total  volume  is  not  more  than  4  or  5  cc. 

If  a  precipitate,  which  is  at  first  uniformly  suspended  in  a  liquid, 
is  allowed  to  settle  in  a  tall  jar  until  it  occupies  but  £  of  the  original 
volume  of  the  mixture  (Fig.  4),  any  soluble  substances  will  still 
remain  uniformly  distributed  throughout  the  whole  volume.  If 
now  the  upper  £ ,  consisting  of  the  clear  solution,  is  drawn  away, 
it  follows  that  practically  £  of  the  solution,  containing  I  of  the 
soluble  impurities,  remains  with  the  precipitate.  By  stirring  up 
the  solid  again  with  pure  water,  the  soluble  impurities  become  uni- 
formly distributed  through  the  larger  volume,  and  on  letting  the 
precipitate  settle  and  drawing  off  |  of  the  liquid,  as  before,  there 
will  remain  with  the  wet  precipitate  only  £  X  i  =  yo  °f  the  origi- 
nal soluble  matter.  After  the  third  decantation  the  remaining 
suspension  will  contain  |  X  -fa  =  yir  of  the  original  impurities, 
and  so  on. 

6.     EVAPORATION 

(a)  When  it  is  necessary  to  remove  a  part  of  the  solvent  from  a 
solution,  as  when  a  dissolved  substance  is  to  be -crystallized  out, 
the  solution  is  evaporated.  In  some  cases,  where  the  dissolved 
substance  is  volatile  or  is  decomposed  by  heat,  the  evaporation 
must  take  place  at  room  temperature,  but  ordinarily  the  liquid 
may  be  boiled.  The  boiling  down  of  a  solution  should  always  be 
carried  out  in  a  porcelain  dish  of  such  a  size  that  at  the  outset  it  is 
well  filled  with  the  liquid.  (Never  evaporate  in  a  beaker.)  The 
flame  should  be  applied  directly  under  the  middle  of  the  dish 
where  the  liquid  is  deepest ;  the  part  of  the  dish  against  which  the 
flame  plays  directly  should  be  protected  with  wire  gauze.  Under 
no  circumstances  should  the  flame  be  allowed  to  play  up  over  the 
sides  of  the  dish:  first,  because,  by  heating  where  the  dish  is  part 
cooled  by  liquid  and  part  uncooled,  there  is  great  danger  of  break- 
ing; second,  because  by  heating  the  sides  the  film  of  liquid  which 
creeps  up  is  evaporated  and  the  solid  deposited  becomes  baked 
hard  and  in  some  cases  is  decomposed.  To  prevent  the  formation 
of  a  solid  crust  around  the  edges,  which  even  at  best  will  take  place 
to  some  extent,  the  dish  should  occasionally  be  tilted  back  and 
forth  a  little,  so  that  the  crust  may  be  dissolved,  or  loosened,  and 
washed  back  into  the  middle  of  the  dish. 

While  evaporating  a  liquid  over  a  flame  it  should  be  carefully 


12 


NOTES  ON  LABORATORY  MANIPULATION 


watched,  for  if  it  should  be  forgotten  and  evaporate  to  dryness  the 
dish  would  probably  break  and  the  product  be  spoiled.  If  a  pre- 
cipitate or  crystals  separate  from  the  liquid  and  collect  in  a  layer 
at  the  bottom,  the  dish  will  probably  break,  because  where  the 
solid  prevents  a  free  circulation  of  the  liquid  the  dish  becomes 
superheated,  and  then  when  in  any  one  place  the  liquid  does  pene- 
trate, the  sudden  cooling  causes  the  porcelain  to  crack.  Usually 
when  a  solid  begins  to  separate  from  a  boiling  liquid  the  evapora- 
tion should  be  stopped  and  the  liquid  left 
to  crystallize.  After  that  the  mother 
liquor  may  be  evaporated  further  in  a 
smaller  dish. 

(b)  Evaporating  to  Dryness.  —  The  only 
circumstances  under  which  a  direct  flame 
may  be  used  to  evaporate  to  dryness  are 
that  the  dish  shall  be  held  in  the  hand 
all  the  time  and  the  contents  rotated  to 
keep  the  sides  of  the  dish  wet. 

Steam  Bath.  —  Laboratories  are  some- 
times equipped  with  general  steam  baths, 
which  are  copper  or  soapstone  chests  kept 
filled  with  steam,  and  provided  with  round 
openings  in  the  top  into  which  evaporat- 
ing dishes  may  be  set. 

Each  student,  however,  may  set  up  a 
steam  bath  as  shown  in  Fig.  5,  at  his  own 
desk.  After  the  water  in  the  beaker  is  once 
boiling  a  very  small  flame  is  all  that  should  be  used,  because  the 
steam  that  escapes  around  the  sides  of  the  dish  is  wasted;  only  as 
much  as  will  condense  on  the  bottom  of  the  dish  is  effective. 
With  such  a  steam  bath  there  is  no  danger  of  spattering  nor  of 
decomposing  the  solid  product  while  evaporating  a  solution  to 
dryness. 

7.    DISSOLVING  SOLID  SUBSTANCES 

The  process  of  dissolving  solid  substances  is  hastened,  first 
by  powdering  the  substance  as  finely  as  possible,  and  second  by 
raising  the  temperature.  The  solid  and  solvent  should  be  heated 
together  in  a  porcelain  dish  (not  in  a  beaker),  and  care  should  be 
taken  to  keep  the  mixture  well  stirred,  for  if  the  solid  should  settle 


FIG.  5.  Steam  Bath  for 
Evaporating  to  Dry- 
ness.  Note  6  (6) 


CRYSTALLIZATION  13 

in  a  layer  on  the  bottom,  that  part  of  the  dish  would  become  super- 
heated and  would  be  apt  to  break  (see  last  paragraph  in  Note  6  (a)). 
The  finer  particles  of  the  solid  dissolve  first;  as  the  solution 
becomes  more  concentrated  the  rate  of  solution  grows  slower,  and 
it  would  take  the  remaining  coarser  particles  a  very  long  time  to 
dissolve.  Hence  when  a  limited  amount  of  solvent  or  reagent 
is  used,  as  for  example  when  copper  is  to  be  dissolved  in  a  minimum 
amount  of  nitric  acid,  it  is  best  to  hold  in  reserve  perhaps  one-tenth 
of  the  reagent ;  when  the  nine-tenths  are  almost  exhausted  and  the 
reaction  with  the  coarser  particles  has  almost  stopped,  pour  off  the 
solution  already  obtained,  and  treat  the  small  residue  with  the 
fresh  acid  held  in  reserve. 


8.    CRYSTALLIZATION 

(a)  A  great  number  of  pure  substances  are  capable  of  assuming 
the  crystalline  condition  when  in  the  solid  form.  Crystals  are 
bounded  by  plane  surfaces,  which  make  definite  and  characteristic 
angles  with  each  other  and  with  the  so-called  axes  of  the  crystals. 

The  external  form  of  a  crystal  reflects  in  some  manner  the  shape 
or  structure  of  the  individual  molecules  of  the  substance;  for  the 
crystal  must  be  regarded  as  being  built  up  by  the  deposition  of 
layer  on  layer  of  molecules,  all  of  which  are  placed  in  the  same 
definite  spatial  relation  to  the  neighboring  molecules. 

When  a  substance  takes  on  the  solid  form  very  rapidly  (as  when 
melted  glass  or  wax  cools)  its  molecules  do  not  have  an  opportunity 
to  arrange  themselves  in  a  regular  order,  and  consequently  the 
solid  body  is  amorphous.  The  axes  of  the  individual  molecules 
point  in  every  direction  without  regularity,  and  consequently  the 
solid  body  possesses  no  crystalline  axes  or  planes. 

It  is  evident  from  the  above  that  the  essential  condition  favor- 
ing the  formation  of  perfect  crystals  is  that  the  solid  shall  be  built 
up  very  slowly.  This  is  the  only  general  rule  which  can  be  given 
in  regard  to  the  formation  of  perfect  crystals. 

The  excellence  of  a  chemical  preparation  is  in  many  cases 
judged  largely  from  its  appearance.  The  more  uniform  and  perfect 
the  crystals,  the  better  appearance  the  preparation  presents. 

In  the  following  preparations  sometimes  a  pure  melted  sub- 
stance is  allowed  to  crystallize  by  simply  cooling;  in  such  a  case 
the  cooling  should  take  place  slowly.  More  often  crystals  are 


14  NOTES  ON  LABORATORY  MANIPULATION 

formed  by  the  separation  of  a  dissolved  substance  from  a  saturated 
solution.  Perfect  crystals  can  best  be  obtained  in  this  case  by 
keeping  the  solution  at  a  constant  temperature  and  allowing  it  to 
evaporate  very  slowly.  This  is  easily  accomplished  in  industrial 
works  where  large  vats  of  solution  can  be  kept  at  a  uniform  tem- 
perature with  steam  coils  and  allowed  to  evaporate  day  and  night. 
On  the  laboratory  scale  it  is  almost  impossible,  first  on  account 
of  variations  in  temperature,  and  next  on  account  of  dust  which 
must  fall  into  an  uncovered  dish. 

The  majority  of  substances  are  more  soluble  at  higher  temper- 
atures than  at  lower.  If  a  solution  just  saturated  at  a  high 
temperature  is  allowed  to  cool  very  slowly,  it  is  possible  for  the 
solid  to  separate  so  slowly  as  to  build  up  perfect  crystals.  This  is 
an  expedient  that  can  be  adopted  to  advantage  in  several  of  the 
preparations.  In  many  cases,  however,  when  a  saturated  solution 
cools  it  becomes  supersaturated,  sometimes  to  a  high  degree.  Then 
when  crystallization  is  once  induced  it  occurs  with  such  rapidity 
that  a  mass  of  minute  crystals,  instead  of  a  few  large,  perfect  ones, 
is  produced.  To  avoid  this  supersaturation  a  few  seed  crystals 
(i.e.,  very  small  crystals  of  the  kind  desired)  may  be  placed  in  the 
solution  before  it  has  cooled  quite  to  the  saturation  point.  These 
form  nuclei  on  which  large  crystals  can  be  built  up,  and  when  they 
are  present  it  is  impossible  for  the  solution  to  remain  super- 
saturated. 

In  carrying  out  the  following  preparations  the  principles  just 
stated  should  be  kept  carefully  in  mind;  but  in  many  instances 
specific  suggestions  will  be  given  as  to  the  easiest  method  for  ob- 
taining good  crystals  of  any  particular  substance. 

Large  crystals,  it  is  true,  present  a  pleasing  appearance,  but 
oftentimes  they  contain  a  considerable  quantity  of  the  mother 
liquor  inclosed  between  their  crystal  layers.  Hence  if  purity  of 
product  is  the  sole  requisite,  it  is  often  more  desirable  to  obtain 
a  meal  of  very  fine  crystals.  Such  a  meal  is  obtained  by  crystal- 
lizing rapidly  and  stirring  while  crystallizing.  Some  substances 
are  so  difficult  to  obtain  in  large  crystals  that  it  is  more  satisfactory 
to  try  only  to  obtain  a  uniform  crystal  meal. 

(6)  Purification  by  Recrystallization.  —  When  a  given  substance 
crystallizes  from  a  solution,  it  most  generally  separates  in  a  pure 
condition  irrespective  of  any  other  dissolved  substances  the  solu- 
tion may  contain.  Thus  a  substance  can  be  obtained  in  an  ap- 


DRYING  15 

proximate  state  of  purity  by  a  single  crystallization.  Portions 
of  the  mother  liquor  (containing  dissolved  impurities)  are,  how- 
ever, usually  entrapped  between  the  layers  of  the  single  crystals, 
not  to  mention  the  liquid  which  adheres  to  the  crystal  surfaces. 
By  dissolving  the  crystals,  the  small  amount  of  impurity  likewise 
passes  into  the  solution,  but  only  a  small  fraction  of  this  impurity 
is  later  entrapped  by  the  crystals  when  they  separate  from  this 
mother  liquor.  By  several  recrystallizations,  then,  a  substance 
can  be  obtained  in  a  very  high  state  of  purity. 

9.    DRYING 

(a)  A  preparation  that  is  not  affected  by  the  atmosphere  can 
be  dried  by  being  spread  in  a  thin  layer,  allowing  the  liquid  adher- 
ing to  the  grains  or  crystals  to  evaporate.     Paper  towels  are 
extremely  useful  in  drying  preparations  because  a  great  deal  of  the 
moisture  is  absorbed  into  the  pores.     When  a  corrosive  liquid,  for 
example  nitric  acid,  clings  to  the  product,  the  latter  is  best  spread 
on  an  unglazed  earthenware  dish,  which  absorbs  the  liquid  without 
being  attacked  by  it.      During  the  drying  the  material  should 
occasionally  be  turned  over  with  a  spatula. 

If  the  material  is  not  decomposed  by  heat  it  can  be  dried  much 
more  rapidly  in  a  warm  place,  as  on  a  steam  heated  iron  plate 
(steam  table);  but  a  product  containing  water  of  crystallization 
should  never  be  dried  at  an  elevated  temperature.  During  the  drying 
the  preparation  must,  of  course,  be  carefully  protected  from  dust. 

(b)  Efflorescent  Crystals  and  Crystals  which  Absorb  Carbon  Di- 
oxide should  be  quickly  pressed  between  paper  towels  until  as 
much  as  possible  of  the  liquid  is  soaked  up,  and  then  they  should 
be  wrapped  in  a  tight  package  in  several  layers  of  fresh  paper 
towels  and  left  in  the  cupboard  24  hours  or  longer.     The  liquid 
is  drawn  by  capillarity  into  the  paper  and  evaporates  from  the 
outer  surface,  but  the  paper  so  impedes  the  circulation  of  air  that 
water  vapor  does  not  escape,  and  the  crystals  will  not  effloresce  in 
several  days.     With  preparations  that  react  with  carbon  dioxide 
(such  as  barium  hydroxide)  the  solution  which  soaks  into  the  paper 
retains  all  of  the  carbon  dioxide  which  might  otherwise  get  in  to 
contaminate  the  product. 

(c)  Substances  which  decompose  on  standing  exposed  to  the  air 
may  be  quickly  dried  if  they  are  first  rinsed  with  alcohol,  or  with 


16       NOTES  ON  LABORATORY  MANIPULATION 

alcohol  and  then  ether.  Rinsing  with  alcohol  removes  nearly  all 
of  the  adhering  water,  and  a  further  rinsing  with  ether  removes 
the  alcohol.  Alcohol  evaporates  more  rapidly  than  water,  but 
ether  evaporates  so  rapidly  that  a  preparation  wet  with  it  may  be 
dried  by  a  very  few  minutes'  exposure  to  the  air. 

Alcohol  and  ether  are  both  expensive  and  should  be  used  spar- 
ingly. They  can  be  used  most  effectively  as  follows:  After  all 
the  water  possible  has  been  drained  from  the  preparation,  transfer 
the  latter  to  an  evaporating  dish  and  pour  over  it  enough  alcohol 
to  thoroughly  moisten  it;  stir  it  with  a  spatula  until  the  alcohol 
has  penetrated  to  every  space  between  the  crystal  grains,  then 
pour  off,  or  drain  off,  the  alcohol  and  treat  the  preparation  in  like 
manner  with  another  portion  of  fresh  alcohol.  After  that  wash 
it  once  or  twice  with  ether  in  exactly  the  same  manner.  If  the 
preparation  is  washed  on  the  filter,  drain  off  the  water  as  thor- 
oughly as  possible,  stop  the  suction,  add  just  enough  alcohol  to 
moisten  the  whole  mass,  and  after  letting  it  stand  a  few  moments 
drain  off  the  liquid  completely.  Apply  a  second  portion  of  alcohol 
and  portions  of  ether  in  the  same  manner. 

10.    PULVERIZING 

In  chemical  reactions  in  which  solid  substances  are  involved  the 
action  is  limited  to  the  surface  of  the  solid,  and  for  this  reason  it  is 
evident  that  it  must  be  much  slower  than  reactions  which  take 
place  between  dissolved  substances;  it  is  also  evident  that  the 
more  finely  powdered  a  solid  substance,  the  greater  is  its  surface, 
and  therefore  the  more  rapidly  it  will  react. 

Most  solid  raw  materials  for  the  following  preparations  are 
supplied  in  the  powdered  form,  but  they  are  rarely  powdered  finely 
enough;  they  should  in  general  be  further  pulverized  until  they 
no  longer  feel  gritty  beneath  the  pestle  or  between  the  fingers. 

For  grinding  any  quantity  of  a  substance  a  large  porcelain 
mortar  (say  8  inches  in  diameter)  with  a  heavy  pestle  is  preferable 
to  the  small  mortars  usually  supplied  in  the  desks.  One  or  more 
such  mortars  is  placed  in  the  laboratory  for  general  use. 

If  a  hard  substance  can  be  obtained  only  in  large  pieces,  it  should 
first  be  broken  with  a  hammer,  then  crushed  into  small  particles 
in  an  iron  or  steel  mortar,  after  which  it  is  to  be  ground  in  the 
porcelain  mortar.  In  the  final  grinding  it  is  often  advisable  to  sift 


DRY  REACTIONS;    FURNACES  17 

the  fairly  fine  from  the  coarser  particles,  then  to  finish  grinding 
the  former  by  itself  and  to  crush  and  grind  the  coarser  particles 
apart 

11.     NEUTRALIZING 

Various  indicators  are  used  to  determine  whether  a  solution  is 
acidic  or  basic.  For  example  litmus  is  red  in  the  presence  of  acid, 
blue  in  the  presence  of  a  base,  and  of  an  intermediate  purple  tint 
in  pure  water  or  a  neutral  salt  solution.  When  a  solution  which 
is  acid  must  be  rendered  exactly  neutral,  base  is  added  until  the 
solution  gives  the  neutral  tint  to  litmus.  If  the  contamination 
will  do  no  harm,  a  drop  or  two  of  the  litmus  solution  is  added 
direct!}*-  to  the  liquid,  otherwise  a  drop  of  the  liquid  must  be  with- 
drawn on  a  stirring  rod  and  touched  to  a  piece  of  litmus  paper. 

It  is  a  tedious  operation  to  exactly  neutralize  a  solution  in  this 
way,  but  the  process  is  greatly  facilitated  if  a  fraction  of  the  liquid 
to  be  neutralized  is  held  in  reserve  in  another  vessel.  The  reagent 
may  be  added  rather  freely  to  the  main  portion  until  the  neutral 
point  is  not  only  reached  but  overstepped.  Then  a  part  of  the 
reserve  may  be  added  and  the  reagent  again  added,  but  more 
cautiously  this  time,  and  so  on  until  the  whole  solution  is  exactly 
neutralized. 

The  procedure  outlined  in  the  last  paragraph  is  a  general  one 
to  follow  whenever  adding  a  reagent  which  must  be  used  in  exactly 
the  right  amount  and  not  in  excess: — always  save  a  fraction  of  the 
original  material  in  reserve  before  adding  the  reagent  to  the  main 
portion. 

12.    DRY  REACTIONS;  FURNACES 

Dry  solid  substances  do  not  react  appreciably  with  each  other 
at  ordinary  temperature.  Reactions  are  made  possible  in  two 
ways :  first,  the  wet  way,  in  which  the  substances  are  dissolved  and 
thus  brought  into  most  intimate  contact.  In  many  cases  solution 
also  produces  ionization,  which,  as  is  known,  greatly  increases 
chemical  activity. 

Reactions  in  the  dry  way  are  rendered  possible  by  heat.  Heat 
alone  increases  the  rapidity  of  a  chemical  reaction,  it  being  a  gen- 
eral law  that  the  speed  is  increased  from  two  to  three  times  for 
every  increase  of  10°  C.  in  temperature.  In  cases  in  which  one 
or  more  of  the  reacting  substances  are  melted  by  the  heat,  the 


18 


NOTES  ON  LABORATORY  MANIPULATION 


same  sort  of  intimate  contact  is  brought  about  as  in  the  case  of 
solutions.  Fusion  is  likewise  a  means  of  producing  electrolytic 
dissociation,  and  on  this  account  also  it  increases  chemical  activity. 
In  some  of  the  furnace  reactions  in  which  none  of  the  substances 
are  melted,  as,  for  Cample,  in  the  reduction  of  strontium  sul- 
phate to  strontium  sulphide  ;by  means  of  charcoal  (see  Prepa- 


FIG.  6 

ration  No.  20),  the  process  probably  takes  place  in  virtue  of  a  cer- 
tain amount  of  gas  which  is  continuously  regenerated.  A  little 
of  the  hot  charcoal  is  oxidized  to  carbon  monoxide,  which  then 
reduces  some  of  the  strontium  sulphate,  it  being  itself  changed 
to  carbon  dioxide  thereby;  the  latter  gas  comes  in  contact  with 
incandescent  charcoal,  and  carbon  monoxide  is  again  produced. 

Reactions  in  the  dry  way  are  usually  carried  out  in  crucibles  — 
of  iron,  clay,  or  graphite,  according  as  to  which  is  least  attacked 
by  the  reagents.  For  rather  moderate  temperatures  the  crucible 
may  be  heated  over  a  flame,  in  other  cases  the  requisite  temper- 
ature can  be  obtained  in  a  furnace. 

The  form  of  furnace  to  be  recommended  for  this  work  is  repre- 
sented in  Figure  6.  It  consists  of  a  cylinder  of  fire  clay,  7  inches 
high  and  6J  inches  in  external  diameter,  which  is  surrounded  by 
a  sheet  iron  casing.  It  is  heated,  as  shown,  by  a  gas-wind  flame, 
introduced  through  an  opening  in  the  lower  part  of  one  side.  If 
a  suitable  air  blast  is  not  available,  a  gasoline  blowpipe  (such  as 
is  commonly  used  by  plumbers)  is  serviceable. 

When  such  a  furnace  as  that  described  is  heated  as  hot  as  possible 
with  a  well-regulated  mixture  of  gas  and  air,  a  temperature  of 
about  1,350°  can  be  obtained.  For  carrying  out  ordinary  chemical 
preparation  work  an  accurate  enough  measure  of  the  temperature 


GAS  GENERATORS 


19 


is  given  by  the  color  of  the  glowing  interior  of  the  furnace,  and 
the  approximate  centigrade  values  corresponding  to  different  colors 
are  as  follows : 


Incipient  red  heat 
Dull  red  heat  . 
Red  heat  .  ;  . 
Bright  red  heat 
Yellow  heat  . 
White  heat 


550° 

650° 

800° 

1,000° 

1,200° 

1,350° 


13.     GAS  GENERATORS 

(a)  Carbon  Dioxide,  Hydrogen,  and  Hydrogen  Sulphide.  —  The 
simplest  form  of  generator  for  these  gases  is  shown  in  Fig.  7. 
The  solid  material,  cracked  marble  for  carbon  dioxide,  feathered 
zinc  for  hydrogen,  and  ferrous  sul- 
phide for  hydrogen  sulphide,  is  placed 
in  the  300  cc.  thick  walled  generator 
bottle.  The  tubes  are  fitted  as  shown 
and  in  the  drying  tube  is  placed  a  plug 
of  cotton  wool  to  strain  the  acid  spray 
out  +.'  the  gas,  or  if  the  gas  is  to  be 
dried,  granulated  calcium  chloride  held 
in  place  with  a  plug  of  cotton  wool  on 
either  side.  Enough  water  is  poured 
in  through  the  thistle  tube  to  cover  its 
lower  end  and  then  about  5  cc.  of  6-n 
HC1.  The  gas  begins  to  generate  rather 
slowly  but  if  one  is  impatient  and  adds 
more  acid  at  once  the  action  will  soon 
become  so  violent  as  to  drive  foam  out 
through  the  delivery  tube.  After  a  few 
minutes  add  more  acid,  1  cc.  at  a  time,  in  order  to  keep  up  the 
evolution  of  gas  at  the  desired  rate. 

(6)  Oxygen  and  Acetylene.  —  The  apparatus  shown  in  Fig.  8  is 
more  suited  for  generating  these  gases,  which  are  produced  by 
allowing  water  to  drip  respectively  on  sodium  peroxide  and 
calcium  carbide.  Remove  the  fittings  from  the  flask  and  place 
in  it  the  required  amount  of  dry  material.  The  sodium  peroxide 
to  be  used  comes  under  the  trade  name  of  "  oxone: "  it  has  been 


FIG. 


20 


NOTES  ON  LABORATORY  MANIPULATION 


fused  and  then  cracked  into  good  sized  lumps  which  are  so  hard 
that  they  do  not  react  with  too  much  violence  with  water.  With 
the  fittings  still  removed  from  the  flask,  fill  the  thistle  tube  with 

water,  open  the  pinch  cock  and 
,  allow  the  vertical  tube  to  become 
completely  filled,  then  close  the 
pinch  cock.  Replace  the  fittings 
in  the  flask.  Open  the  pinch  cock 
cautiously  to  let  a  single  drop  of 
water  fall  on  the  material.  Note 
the  effect  and  thereafter  let  the 
water  in,  a  single  drop  at  a  time, 
to  obtain  the  desired  flow  of  gas. 

(c)  Automatic  Gas  Generator  for 
Carbon  Dioxide,  Hydrogen  and 
Hydrogen  Sulphide.  —  The  appara- 
tus shown  in  Fig.  9  is  based  on 
the  principle  of  the  familiar  Kipp 
generator  and  it  is  especially 
suited  to  cases  in  which  a  solution 
is  to  be  saturated  with  the  gas  in 
question,  as,  for  example,  when  an 
ammoniacal  solution  of  common 
salt  is  to  be  saturated  with  carbon 

dioxide  in  the  preparation  of  sodium  bicarbonate  by  the  Solvay 
process. 

Assemble  the  apparatus  as  shown  in  the  diagram.  The  stem 
of  the  generator  tube  E  should  reach  flush  with  the  bottom  of  the 
stopper  but  not  below.  The  delivery  tube  C  should  reach  nearly 
to  the  bottom  of  the  generator  bottle  D.  Place  the  requisite 
amount  of  calcium  carbonate  (or  zinc,  or  ferrous  sulphide)  in  the 
generator  tube.  Then  insert  a  loose  plug  of  glass  wool  F  about 
li  inches  long  so  that  it  will  stand  about  midway  between  the 
top  of  the  solid  material  and  the  stopper  in  the  mouth  of  the  tube, 
and  act  as  a  gas  filter  (to  remove  acid  spray).  Pour  the  requisite 
amount  of  acid  into  the  reservoir  A ;  clamp  the  reservoir  at  just 
the  same  height  as  the  generator  tube  and  pour  in  water  cautiously 
until  the  acid  rises  and  barely  touches  the  solid  in  the  generator 
tube.  The  generation  of  gas  will  now  begin  and  proceed  auto- 


FIG.  8 


GAS  GENERATORS 


21 


matically  as  fast  as  the  gas  is  allowed  to  flow  from  the  delivery 
tube,  H. 

Place  the  solution  to  be  saturated  with  the  gas  in  the  flask  G, 
insert  the  stopper  and  delivery  tube,  but  let  the  stopper  remain 
loose  until  the  air  is  entirely  expelled.  Then  make  the  stopper 


FIG.  9 

tight;  the  gas  will  pass  in  as  rapidly  as  it  can  be  absorbed  by  the 
solution.  Shaking  the  receiving  flask,  will  greatly  increase  the 
rapidity  of  absorption,  but  this  should  be  done  with  a  good  deal 
of  caution  at  first,  because  if  the  undiluted  acid  is  drawn  too  sud- 
denly up  in  the  tube  E  the  violence  of  the  reaction  may  either  blow 
out  the  stoppers  or  drive  foam  through  the  glass  wool  filter  F  and 
into  the  delivery  tube  H.  After  the  solution  is  partly  saturated 
the  flask  may  be  shaken  continuously  and  the  reservoir  A  raised 
to  a  higher  level. 


CHAPTER  I 
THE  QUANTITATIVE  ASPECTS   OF  CHEMISTRY 

So  many  things  happen  when  substances  undergo  a  chemical 
change  that  it  is  no  wonder  the  student  is  astonished  and  even 
bewildered  by  his  observations.  He  soon  learns  that  there  is  an 
entire  change  in  physical  properties  of  the  reacting  substances  and 
finds  that  this  phenomenon  is  usually  the  most  easy  to  observe. 
Also  he  is  able  to  discover  the  transformations  of  energy  which 
always  accompany  a  chemical  change,  although  this  is  usually 
confined  to  observations  of  the  evolution  of  heat. 

He  learns,  usually  from  the  text  book,  that  no  mass  is  lost  or 
gained  during  a  chemical  change  and  that  the  total  weight  of  the 
substances  before  and  after  is  the  same.  This  rule  is  known 
as  the  law  of  the  conservation  of  matter,  and  it  is  more  difficult 
for  the  student  to  convince  himself  of  the  truth  of  this  character- 
istic of  chemical  change  by  direct  observation  because  the  ex- 
periments must  be  quantitative  and  require  apparatus  for  meas- 
uring and  weighing. 

When  hydrogen  combines  with  oxygen  to  form  water  it  is  easily 
observed  that  a  great  deal  of  heat  is  given  off,  also  that  a  liquid 
substance  is  formed  and  that  the  gases  taken  decrease  in  volume 
and  in  fact  entirely  disappear  when  exactly  the  right  mixture  is 
used.  When  measuring  tubes  are  used  it  is  observed  that  when 
two  volumes  of  hydrogen  and  one  volume  of  oxygen  are  mixed, 
the  gases  entirely  disappear  after  the  reaction  has  taken  place. 
When  more  than  two  volumes  of  hydrogen  are  taken  to  one  of 
oxygen,  the  excess  of  hydrogen  over  the  two  volumes  is  found  to 
remain  unaffected;  and  likewise  when  more  than  one  volume  of 
oxygen  is  taken  to  two  volumes  of  hydrogen,  the  excess  of  oxygen 
over  the  one  volume  is  found  to  remain  unchanged  after  the  re- 
action. But  to  learn  that  1.008  parts  by  weight  of  hydrogen  com- 
bine with  exactly  8.00  parts  by  weight  of  oxygen  requires  elabo- 
rate apparatus  and  very  painstaking  measurements.  This  pro- 
portion, 1.008  :  8.00,  when  determined,  is  known  as  the  combining 
ratio. 

22 


EXPERIMENTS  23 

Similarly,  exact  ratios  hold  with  respect  to  other  chemical 
changes.  When  zinc  and  oxygen  combine  to  form  zinc  oxide 
the  ratio  is  found  to  be  one  of  oxygen  to  4.09  of  zinc,  no  matter 
how  much  zinc  or  oxygen  is  used.  The  excess  of  one  or  the 
other  is  left  over  unless  they  are  used  in  just  this  proportion. 

Some  substances  do  not  combine  with  each  other  but  each  may 
combine  with  a  third.  Under  such  conditions  a  ratio  may  be 
calculated  between  the  two  substances  that  did  not  combine. 


Experiments 
WEIGHING 
Two  kinds  of  apparatus  may  be  obtained  for  weighing: 

1.  Rough  balances  or  platform  scales.    These  are  to  be  used 
for  weighing  out  materials  approximately  for  qualitative  experi- 
ments and  for  weighing  heavy  objects  of  over  100  grams. 

2.  Sensitive  balances,  which  will  weigh  accurately  to  a  centi- 
gram.    These  are  the  balances  in  the  glass  cases.     They  are  not 
as  sensitive  as  the  best  analytical  balances,  which  weigh  to  one- 
tenth  of  a  milligram,  but  they  are  sufficiently  accurate  to  do  some 
kinds  of  work  extremely  well  and  are  so  sensitive  that  they  require 
careful  and  intelligent  handling.     Hence  before  using  one  of  these 
balances  apply  to  an  instructor  for  individual  instruction  as  to  its 
manipulation.     The  balances  must  not  be  used  until  permission 
is  obtained.    The  following  general  rules  must  always  be  ob- 
served: 

1.  No  load  of  over  100  grams  must  be  put  on  the  sensitive 
balances.     By  no  chance  will  any  heavier  object  than  this  need  to 
be  weighed  with  a  greater  precision  than  can  be  obtained  on  the 
^platform  scales. 

2.  The  material  to  be  weighed,  unless  it  is  in  a  single,  clean, 
dry  piece,  should  never  be  placed  directly  on  the  scale  pan.     If 
powdery,  place  on  a  piece  of  paper,  if  wet  on  a  watch  glass  or  in  a 
beaker,  the  outside  of  which  is  clean  and  dry. 

3.  To  be  accurate  the  weights  must  be  kept  clean.     Never 
touch  them  with  the  hands,  but  use  the  forceps. 

4.  When  altering  load  or  weights,  the  scale  pans  must  rest  on 
the  floor  of  the  balance  case.     Never  leave  the  lever  raised  ex- 
cept in  taking  the  final  swing  after  the  weights  are  adjusted. 


24  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 

During  the  swing  the  balance  door  must  be  closed,  and  oefore  it 
is  opened  the  pans  must  again  be  lowered. 

5.  The  same  balance  and  set  of  weights  must  be  used  through- 
out any  one  experiment  in  order  that  possible  errors  in  the  weights 
or  balances  may  cancel  in  the  successive  weighings. 

1.    THE   COMBINING  RATIO  OF  ZINC  AND  OXYGEN 

Zinc  and  zinc  oxide  are  both  substances  whose  weight  can  be 
accurately  determined,  so  that  the  quantity  of  oxygen  which  is 
combined  in  the  oxide  can  be  found  by  difference.  To  convert 
quantitatively  a  definite  amount  of  zinc  into  the  oxide  by  means 
of  direct  combination  with  oxygen  would  be  a  difficult  operation; 
but  the  same  result  is  accomplished  indirectly  by  first  treating 
the  metal  with  nitric  acid  to  obtain  the  nitrate  and  then  decom- 
posing the  zinc  nitrate  by  heat,  which  leaves  a  residue  of  zinc 
oxide. 

Weigh  a  4-inch  evaporating  dish  accurately.  Place  in  it  about  1 
gram  of  turnings  of  chemically  pure  zinc  and  again  weigh  accurately. 
Cover  the  dish  with  a  5-inch  watch  glass;  add  5  cc.  of  water  and 
then  add  dilute  nitric  acid,  a  cubic  centimeter  at  a  time,  until 
all  the  zinc  has  dissolved.  Then  remove  the  watch  glass,  and 
with  the  jet  of  the  wash  bottle  rinse  the  drops  clinging  to  it  back 
into  the  dish.  Place  the  solution  on  a  water  bath  (beaker  of 
boiling  water)  and  leave  it  to  evaporate  (with  the  watch  glass 
removed)  as  much  as  possible.  When  only  a  small  amount  of 
sirupy  liquid  (melted  zinc  nitrate)  is  left,  take  the  dish  to  the  hood, 
support  it  on  a  wire  triangle,  and  heat  carefully  with  a  very  small 
flame,  holding  the  burner  in  the  hand.  If  the  liquid  starts  to 
boil,  remove  the  flame  at  once,  because  every  tiny  drop  that 
spatters  out  of  the  dish  means  a  loss  of  material.  Heat  until  the 
mass  thickens  and  red  fumes  escape,  and  finally,  after  the  mass 
grows  perfectly  dry,  heat  quite  strongly  for  a  few  minutes.  Cool 
and  weigh.  Again  heat  quite  strongly  and  weigh.  The  weight 
ought  not  to  have  decreased,  but  if  it  has  the  heating  must  be 
continued  until  it  ceases  to  do  so. 

Calculation.  From  the  weight  of  the  zinc  and  the  zinc  oxide 
find  the  combining  ratio  of  zinc  and  oxygen. 


EXPERIMENTS 


25 


2.    WEIGHT  OF  A  LITER  OF   OXYGEN 

The  volume  occupied  by  a  given  quantity  of  gas  is  influenced 
very  appreciably  by  three  factors:  temperature,  pressure,  and 
content  of  aqueous  vapor.  The  volume  measured  in  this  experi- 
ment must  be  reduced  to  what  is  known  as  standard  conditions, 
that  is,  dry  gas  at  0°  and  760  mm.  pressure  by  means  of  the 
formula  given  on  page  37.  The  oxygen  used  is  obtained  from 
potassium  chlorate  by  heating,  and  its  quantity  is  given  by  the 
loss  in  weight  of  that  material. 

For  the  apparatus  to  measure  the  volume  of  oxygen  evolved 
(see  Figure  10),  clamp  the  wash  bottle  in  an  inverted  position,  re- 


FIG.  10 

move  the  capillary  jet,  and  connect  it  to  the  other  tube  of  the  wash 
bottle  by  means  of  1J  feet  of  rubber  tube,  at  the  middle  of  which 
is  placed  a  pinchcock.  Connect  the  other  tube  of  the  wash  bottle 
by  means  of  an  elbow  tube  with  a  hard  glass  test  tube,  which  is  to 
contain  the  potassium  chlorate.  Test  the  whole  apparatus  and 
make  sure  that  it  is  tight.  Fill  the  wash  bottle  with  water  nearly 
to  the  top  of  the  vertical  tube,  allow  enough  to  run  out  to  fill  the 
rubber  tube  and  jet,  and  close  the  pinchcock. 

Dry  and  weigh  accurately  the  hard  glass  tube.  Place  in  it 
about  2  grams  of  dry,  powdered  potassium  chlorate  and  again 
weigh  accurately.  The  potassium  chlorate  in  the  stock  bottle  is 
not  dry  and  must  not  be  used  for  this  experiment.  Use  that  in  the 


26  QUANTITATIVE  ASPECTS  OF   CHEMISTRY 

bottle  marked  for  this  purpose.  Connect  the  tube  with  the  appa- 
ratus. Dip  the  capillary  jet  in  a  beaker  of  water,  and  raise  the 
latter  until  the  surface  of  the  water  is  at  the  same  level  in  the 
beaker  and  in  the  flask.  Open  the  cock  until  water  has  run  in  or 
out  to  equalize  the  pressure.  Then  close  the  cock.  In  this  way 
the  air  in  the  apparatus  at  tne  start  is  at  atmospheric  pressure. 
At  the  end  it  must  be  brought  to  the  same  pressure,  so  that  the 
two  volumes  are  directly  comparable.  Note:  It  is  important 
that  the  flexible  rubber  tube  be  filled  completely  with  water; 
otherwise,  when  the  levels  are  equalized,  there  will  be  a  column 
of  air  in  one  part  and  a  column  of  water  in  another  part  of  the 
tube  and  the  pressure  on  the  gas  in  the  flask  will  not  be  the  same 
as  that  of  the  atmosphere  on  the  water  in  the  beaker.  Empty 
the  beaker.  Hold  the  jet  pointing  into  the  beaker  so  that  the 
water  may  be  seen  dripping  into  it,  and  thus  the  rate  of  the  pro- 
duction of  gas  regulated.  Open  the  clip  and  commence  heating 
the  potassium  chlorate,  so  that  a  steady,  but  not  a  rapid,  stream 
of  water  runs  from  the  jet.  If  for  any  reason  the  heating  is  in- 
terrupted during  the  process,  submerge  the  jet  beneath  the  liquid, 
so  that  water,  not  air,  shall  be  sucked  back  into  the  tube  and  flask. 
When  about  400  cc.  of  water  have  been  forced  over,  submerge  the 
jet  and  allow  the  ignition  tube  to  cool.  Raise  the  beaker  to 
equalize  the  water  levels  and  close  the  cock.  Record  the  tem- 
perature of  the  laboratory  and  the  barometer  reading.  Weigh 
the  water  in  the  beaker  on  the  platform  scales  (be  sure  not  to  use 
balances  at  this  point),  which  gives  the  actual  volume  of  the  oxy- 
gen evolved.  Calculate  its  volume  under  standard  conditions. 
Weigh  the  tube  again,  and  the  loss  gives  the  weight  of  oxygen. 

Calculation.     Calculate  the  weight  of  1  liter  of  oxygen  under 
standard  conditions. 


3.    VOLUME   OF  HYDROGEN   DISPLACED  BY    ZINC 

By  allowing  an  acid  to  act  upon  zinc  a  definite  weight  of  the 
latter  is  dissolved,  and  the  hydrogen  set  free  can  be  measured  and 
its  volume  corrected  in  a  manner  similar  to  that  of  the  last  ex- 
periment. 

Experiment.  (See  Figure  11).  Clamp  a  good  sized  drying  tube 
in  a  vertical  position  with  its  larger  end  down.  Fit  this  end  with 
a  stopper  through  which  passes  a  short  glass  tube.  Connect  the 


EXPERIMENTS 


27 


latter  by  means  of  1£  feet  of  rubber  tube,  provided  with  a  pinch- 
cock,  with  a  funnel  supported  upright  in  a  considerably  higher 
position  than  the  drying  tube.  From  the  other  end  of  the  dry- 
ing tube  lead  a  1J  foot  rubber  tube  to  a  trough.  Over  the  end 
of  this  tube,  which  should  have  a  right  angle,  3-inch  glass  tube, 


Drying-Tube 
Burette  Clamp 
Rubber  Tube 


Small  Flask 


Large  Pan 


Glass  .Elbow        Small  Pan 

FIG.  11 

invert  a  250  cc.  flask  completely  filled  with  water.  Inside  the 
drying  tube  is  to  be  placed  the  zinc.  Obtain  a  rod  of  it  about 
1 J  inches  long  and  J  inch  in  diameter.  Clean  and  dry  it  and  weigh 
it  accurately.  A  short  piece  of  twisted  copper  wire  supports  the 
zinc  in  the  tube.  Pour  water  into  the  funnel  until  the  whole  ap- 
paratus is  filled  and  close  the  pinchcock.  When  everything  is 
ready  pour  hydrochloric  acid  into  the  funnel  and  allow  it  to  pass 
the  pinchcock  until  hydrogen  is  evolved  from  the  metal.  The 
cone  of  the  funnel  should  always  be  kept  well  filled  with  liquid, 
and  care  must  be  taken  that  no  bubble  of  air  is  sucked  into  the 
stem.  Collect  the  hydrogen  in  the  flask,  and  when  the  gas  has 
forced  the  level  of  the  liquid  nearly  into  the  neck  begin  to  wash  the 
acid  out  of  the  apparatus  by  pouring  water  through.  Have  the 
acid  completely  removed,  and  thus  the  evolution  of  hydrogen 
stopped  by  the  time  the  water  level  stands  at  about  the  middle  of 
the  neck.  Equalize  the  level  inside  and  outside  the  neck  of  the 
flask,  and  while  in  this  position  stopper  it  or  mark  the  level. 


28  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 

Record  temperature  and  barometric  reading.  Determine  on  the 
platform  scales  the  weight  of  water  required  to  fill  the  volume 
occupied  by  the  hydrogen.  Dry  the  zinc  and  weigh  it  again  on 
the  sensitive  balances. 

Calculation.  (1)  Calculate  tjie  volume  of  hydrogen  under  stand- 
ard conditions  equivalent  to  1  gram  of  zinc.  (2)  From  the  com- 
bining ratio  of  zinc  and  oxygen  (Experiment  1),  find  the  volume 
of  hydrogen  equivalent  to  1  gram  of  oxygen.  (3)  From  the  weight 
of  1  liter  of  oxygen  (Experiment  2),  find  the  volume  of  hydrogen 
equivalent  to  1  liter  of  oxygen. 

4.    THE   COMBINING   RATIO  OF  HYDROGEN  AND 
OXYGEN  IN  WATER 

It  is  only  with  great  care  and  refined  apparatus  that  bodies  of 
gases  can  be  successfully  weighed.  It  not  being  feasible  then 
to  weigh  hydrogen  and  oxygen  as  such  in  this  experiment,  the 
oxygen  will  be  obtained  from  solid  copper  oxide,  the  loss  of  weight 
of  which  can  be  determined.  If  an  excess  of  hydrogen  is  led  over 
the  copper  oxide,  all  which  does  not  combine  with  the  oxygen  to 
form  water  will  pass  on  unchanged.  By  retaining  all  the  water 
formed  in  some  material  which  absorbs  it,  its  weight  may  be  found. 
The  amount  of  the  hydrogen  combined  is  then  given  by  the  differ- 
ence between  the  weight  of  the  water  and  that  of  the  oxygen. 

Hydrogen  gas  generated  from  zinc  and  hydrochloric  acid  is 
passed  through  a  tube  containing  calcium  chloride  to  remove  any 
water  vapor,  then  over  copper  oxide,  with  the  oxygen  of  which  it 
combines,  and  then  through  another  calcium  chloride  tube  to 
absorb  the  water  vapor  formed. 

Experiment.  (See  Figure  12.)  To  drive  off  any  water  which  it 
may  contain,  heat  5  to  6  grams  of  copper  oxide  to  redness  in  a 
porcelain  boat.  Place  about  20  grams  of  feathered  zinc  in  a 
generator  bottle.  Through  a  tight  fitting,  two-holed  rubber  stop- 
per pass  a  thistle  tube  reaching  to  the  bottom  and  an  elbow  tube 
just  entering  the  top  of  the  bottle.  To  the  elbow  tube  connect  a 
drying  tube  containing  some  loosely  packed  cotton;  to  this  con- 
nect a  U-tube  with  side  arms  filled  with  granular  calcium  chloride. 
This  passes  into  a  combustion  tube  about  10  inches  long,  which 
should  be  sloped  downward  at  an  angle  of  5°  to  10°.  One  end  of 
the  combustion  tube  should  be  drawn  down  in  the  blast  lamp  to  fit 


EXPERIMENTS 


29 


a  rubber  connector  by  which  it  is  joined  to  the  second  drying  tube. 
Do  not  neglect  to  round  (with  a  file,  or  in  the  flame  of  a  blast 
lamp)  the  sharp  edges  of  the  combustion  tube  before  fitting  the 
rubber  stopper,  as  otherwise  the  latter  would  be  cut  and  it  would 
be  difficult  to  make  a  tight  joint.  The  second  calcium  chloride 


~  Cotton 

alcium  Chloride 
Side  Arm  U-Tub« 


FIG.  12 

tube  should  be  prepared  with  especial  care  and  be  weighed  just 
before  starting  the  operation.  Prolong  its  open  arm  with  an  elbow 
tube  whose  one  end,  drawn  out  to  a  capillary,  is  turned  upward. 
Test  the  tightness  of  the  apparatus :  This  may  be  done  by  filling 
the  generator  bottle  about  J  inch  deep  with  water;  then  holding  a 
finger  on  the  outer  end  of  the  apparatus  and  pouring  more  water 
into  the  thistle  tube.  If  it  sinks  the  apparatus  is  not  tight.  It 
must  be  made  so. 

Add  a  few  cubic  centimeters  of  hydrochloric  acid  to  the  gen- 
erator bottle  and  then  let  hydrogen  generate  and  slowly  fill  the 
apparatus  while  preparing  the  rest  of  the  material.  After  it  has 
cooled  weigh  the  porcelain  boat  filled  with  the  copper  oxide. 
Place  the  boat  carefully,  without  spilling  any  of  the  copper  oxide, 
in  the  middle  of  the  combustion  tube.  Add  a  little  more  acid  to 
the  generator,  and  wait  until  the  apparatus  is  completely  filled 
with  hydrogen  before  proceeding  further.  Never  bring  a  flame 
near  the  apparatus  until  the  purity  of  the  hydrogen  escaping  has 
been  proved.  Test  the  hydrogen  by  collecting  a  3-inch  tube  full 
by  upward  air  displacement  and  carrying  the  tube,  mouth  down- 
ward, to  a  distant  flame.  If  the  gas  does  not  explode,  but  burns 


30  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 

quietly,  convey  the  tube,  still  with  mouth  downward,  to  the  jet. 
If  the  hydrogen  still  burning  in  the  tube  ignites  the  jet,  it  is  then 
safe  to  proceed  with  the  experiment.  The  gas  at  the  jet  may  be 
left  burning.  Increase  the  evolution  of  hydrogen  a  little,  being 
careful  in  adding  acid  to  the  generator  not  to  allow  it  to  drag  with 
it  bubbles  of  air,  as  this  would  introduce  oxygen,  which  would 
be  burned  to  water  in  the  combustion  tube.  Let  the  hydrogen 
generate  5  minutes  longer  to  sweep  any  last  traces  of  oxygen  from 
the  apparatus,  and  then  begin  heating  the  copper  oxide  rather 
gently,  using  the  flame  spreader.  The  water  vapor  formed  will 
partly  condense,  and  great  care  must  be  exercised  that  it  does  not 
crack  the  heated  tube.  Before  the  experiment  is  ended  this 
condensed  water  must  be  completely  driven  over  into  the  calcium 
chloride  tube  by  playing  the  flame  gently  over  the  parts  of  the 
tube  where  moisture  is  seen.  When  the  tube  has  cooled,  discon- 
nect between  the  generator  and  the  drying  tube  and  blow  slowly 
enough  air  (from  the  lungs)  through  the  apparatus  to  displace 
all  the  hydrogen.  Weigh  the  boat  to  find  the  loss,  and  the  cal- 
cium chloride  tube  to  find  the  increase  in  weight. 

Calculation.  Calculate  the  combining  ratio  of  hydrogen  and 
oxygen,  that  is,  how  many  parts  by  weight  of  oxygen  combine 
with  1  part  of  hydrogen. 

From  this  ratio  and  the  volume  ratio  calculated  in  Experiment 
3,  find  the  weight  of  1  liter  of  hydrogen.  This  involves  the 
assumption  that  the  quantity  of  hydrogen  displaced  by  that 
weight  of  zinc  which  combines  with  a  given  amount  of  oxygen  is 
the  same  as  would  combine  directly  with  that  amount  of  oxygen. 
By  comparing  the  result  obtained  with  the  known  weight  of  a 
liter  of  hydrogen  ( =  0.090  gram)  decide  whether  this  assumption  is 
correct. 

Notes  and  Problems 

The  law  of  definite  proportions  states  that  whenever  two  (or 
more)  elements  combine  to  form  a  definite  compound  the  ratio 
by  weight  of  the  elements  entering  that  compound  is  always  the 
same. 

The  law  of  multiple  proportions  comes  within  the  statement  of 
the  law  of  definite  proportions,  but  it  takes  up  the  special  case 
that  two  elements  can  form  more  than  one  definite  compound. 
The  first  law  holds  for  each  of  the  compounds,  the  other  states 


NOTES  AND  PROBLEMS  31 

that  the  definite  ratios  for  the  separate  compounds  are  to  each 
other  in  the  ratio  of  small  whole  numbers.  Perhaps  the  most 
easily  understood  statement  of  the  law  of  multiple  proportions  is 
as  follows:  If  the  same  weight  of  the  first  element  is  taken  in 
each  case,  and  such  weights  of  the  second  element  as  will  combine 
in  each  case  with  the  same  weight  of  the  first  element  to  form  the 
different  compounds  are  taken,  these  weights  of  the  second  element 
are  to  each  other  in  the  ratio  of  simple  whole  numbers. 

The  Atomic  Theory.  Dalton  was  impressed  by  the  signifi- 
cance of  these  two  laws  and  he  found  for  them  a  reasonable  ex- 
planation in  the  atomic  theory  (1808).  According  to  this  theory 
the  elements  consist  of  atoms  which  were  thought  by  Dalton 
to  be  indivisible.  The  atoms  of  the  same  element  are  all  alike 
in  weight  and  all  of  their  other  properties;  the  atoms  of  different 
elements  differ  in  properties.  When  elements  combine,  it  is  the 
individual  atoms  which  are  concerned;  they  attach  themselves  to 
each  other  in  definite  and  very  simple  groupings. 

Suppose  for  example  one  atom  of  element  A  combines  with 
one  atom  of  element  B,  to  form  the  compound  AB,  then,  since 
the  weights  of  atoms  of  the  same  element  are  always  alike  and 
since  whatever  amount  of  the  compound  is  taken  it  always  contains 
an  equal  number  of  atoms  of  each  element,  the  proportion  by 
weight  of  the  two  elements  must  always  be  the  same  in  this 
compound.  Thus  the  law  of  definite  proportions  is  a  necessary 
deduction  from  the  atomic  theory.  Let  it  be  understood  how- 
ever that  the  law  is  a  certain  fact  established  by  careful  measure- 
ments. The  theory  is  simply  the  best  effort  of  the  human 
mind  to  furnish  an  explanation  of  the  facts. 

It  is  likewise  with  the  law  of  multiple  proportions.  Suppose 
that  one  atom  of  A  can  combine  with  two  atoms  of  B  in  forming 
an  entirely  different,  but  none  the  less  definite  compound.  Let 
us  designate  this  compound  AB2.  The  law  of  definite  proportions 
would  hold  for  this  compound  as  well  as  for  the  first.  Further- 
more if  we  should  take  such  amounts  of  each  compound  that  each 
contained  the  same  number  of  atoms  of  A  then  the  second  would 
contain  twice  as  many  atoms  of  B  as  the  first.  The  weight  of  B 
in  the  second  would  therefore  be  exactly  twice  the  weight  of  B  in 
the  first  compound.  Thus  the  law  of  multiple  proportion  is  also 
a  necessary  deduction  from  the  atomic  theory. 


32  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 

Atomic  Weights.  We  have  found  that  the  combining  ratio  of 
oxygen  and  zinc  is  1  :  4.09.  The  atomic  theory  stipulates  that 
the  ratio  of  the  numbers  of  atoms  is  simple,  but  it  may  be  1  :  1, 
1  :  2,  1  :  3;  2  :  1,  3  :  1,  3  :  2,  2  :  3  or  any  reasonably  simple  ratio. 
Having  at  hand  no  way  of  telling  what  the  real  ratio  is  we  pro- 
ceed to  assume  one  to  be  the  "correct  ratio,  —  in  plain  English, 
we  make  a  guess.  Where  we  have  nothing  else  to  guide  us  we 
make  the  simplest  possible  guess  and  assume  the  ratio  to  be 
1:1.  If  this  is  correct  then  the  atom  of  zinc  weighs  4.09  times 
as  much  as  the  atom  of  oxygen. 

The  weight  of  a  single  atom  is  so  very  small  that  it  would  mean 
little  to  us  if  we  did  express  it  exactly  in  arithmetical  figures. 
What  we  are  interested  in  is  the  ratio  of  the  weights.  When  a 
great  many  million  million  atoms  of  oxygen  combine  with  the 
same  number  of  million  million  atoms  of  zinc  we  make  the  actual 
measurement  that  1  gram  of  oxygen  combines  with  4.09  grams  of 
zinc. 

Standard  of  Atomic  Weights,  O  =  16.  It  has  been  inter- 
nationally agreed  that  it  is  a  matter  of  convenience  to  adopt  the 
exact  number  16.000  as  the  atomic  weight  of  oxygen.  Exactly 
16  grams  is  the  gram  atomic  weight  of  oxygen.  We  may  mention 
as  a  matter  of  interest  that  16  grams  of  oxygen  contains  6.06  X 
1023  atoms  although  this  enormous  figure  is  of  no  practical  im- 
portance beyond  the  mere  fact  that  it  is  enormous. 

The  atomic  weight  of  zinc  is  therefore  4.09  X  16  =  65.4  if 
our  assumption  of  the  1  :  1  ratio  is  correct.  65.4  grams  of  zinc 
is  therefore  the  gram  atomic  weight  of  this  element  and  it  likewise 
contains  6.06  X  1023  actual  atoms. 

Of  course  the  fact  that  many  cases  are  known  where  two  ele- 
ments can  combine  in  different  proportions  to  form  different 
compounds  (multiple  proportions)  shows  us  at  once  that  the  atomic 
combining  ratio  cannot  always  be  1  :  1.  In  order  to  establish 
consistent  atomic  weights  for  all  the  elements  on  the  O  =  16 
basis,  we  must  either  make  very  clever  guesses  as  to  the  atomic 
ratios  or  we  must  have  some  reliable  means  of  finding  out  this 
ratio.  We  shall  state  here  that  there  are  reliable  methods  of 
doing  this,  one  of  the  most  useful  of  which  will  be  explained  in  a 
later  section  of  this  chapter.  In  the  front  inside  cover  of  this 
book  is  printed  a  list  of  all  the  elements  with  their  symbols  and 
their  atomic  weights.  The  atomic  weights  are  obtained  in  many 


NOTES  AND  PROBLEMS  33 

cases  from  the  combining  ratio  by  weight  with  oxygen  itself,  — 
in  other  cases  from  the  combining  ratio  with  another  element  whose 
combining  ratio  with  oxygen  is  known.  In  all  cases  the  combining 
ratio  by  weight  with  oxygen  must  be  multiplied  by  16  and  divided 
by  the  atomic  ratio  of  the  element  to  oxygen. 

PROBLEMS 

1.  The  combining  ratio  by  weight  of  zinc  and  sulphur  is 
2.039  :  1.     Assuming  the  knowledge  that  the  atomic  weight 
of  zinc  is  65.4  and  that  zinc  and  sulphur  combine  in  the 
1  :  1  atomic  ratio  find  the  atomic  weight  of  sulphur. 

2.  The  oxide  formed  when  1  gram  of  sulphur  is  burned 
weighs  exactly  2  grams.     What  is  the  combining  ratio  by 
weight  of  sulphur  and  oxygen? 

3.  Using  the  data  and  assumptions  of  the  last  two  prob- 
lems, deduce  the  atomic  ratio  by  which  sulphur  and  oxygen 
combine. 

4.  From  the  atomic  weights  printed  in  the  table,  find  the 
combining  ratio  by  weight  in  the  compounds   having   the 
atomic  ratio  expressed  by  the  formulas  CaO,  Li2O,  Fe203, 
Fe3O4.     Also  find  the  percent  composition  by  weight  of  these 
compounds. 

5.  One  oxide  of  chromium  contains  52  percent  of  chro- 
mium and  48  percent  of  oxygen  by  weight;  another  contains 
68.42   percent  of   chromium   and  31.58  percent  of  oxygen. 
Prove  that  these  compounds  are  in  agreement  with  the  law 
of  multiple  proportions. 

6.  Look  up  the  atomic  weight  of  chromium  and  find  the 
atomic  ratio  in  each  oxide.     Write  the  simplest  formula  of 
each  oxide. 

7.  One  gram  of  potassium  metal  burns  to  give  1.82  grams 
of  an  oxide.     Calculate  the  chemical  formula  of  the  oxide. 

8.  The  oxide  obtained  when  iron  is  burned  in  oxygen  has 
the  composition  Fe  =  72.4;   O  =  27.6.     Calculate  the  chem- 
ical formula  of  the  oxide. 

9.  Find  the  formula  of  the  substance  whose  composition 
is  magnesium  25.57  percent,  chlorine  74.43  percent. 

10.  Find  the  formula  of  the  substance  whose  composition 
is  potassium  26.585,  chromium  35.390,  oxygen  38.025. 


34  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 

Measurement  of  Gases.  Since  it  is  difficult  to  weigh  a  body 
of  gas,  but  comparatively  easy  to  find  its  volume,  the  amounts  of 
gases  are  almost  invariably  estimated  by  measuring  the  volume. 
But  the  volume  of  a  definite  amount  of  gas  is  very  dependent  on 
the  conditions  and  .to  make  a  volume  measurement  have  an 
accurate  meaning  it  becomes  necessary  to  know  exactly  the 
conditions  of  pressure,  temperature,  and  dryness  under  which 
the  measurement  was  made.  To  make  the  results  of  all  measure- 
ments comparable  it  is  customary  to  calculate  what  the  measured 
volume  would  become  if  so-called  standard  conditions  prevailed. 
By  common  consent  of  scientific  men,  standard  conditions  have 
been  defined  as  760  mm.  pressure,  zero  degrees  Centigrade,  and 
dry  gas. 

A  remarkable  uniformity  has  been  found  to  exist  in  the  behavior 
of  all  gases  under  changing  conditions.  Three  simple  state- 
ments, the  so-called  gas  laws  suffice  to  define  with  a  considerable 
degree  of  accuracy  the  volume  changes  with  changing  pressure, 
temperature,  and  water  vapor  content. 

Boyle's  Law  states  that  at  the  same  temperature  the  volume 
of  a  definite  amount  of  a  gas  is  inversely  proportional  to  the  pres- 
sure 


V*         Pi 

or 


Charles'  Law  defines  the  change  of  volume  with  changing 
temperature:  The  volume  of  a  definite  amount  of  a  gas  under 
constant  pressure  is  directly  proportional  to  the  absolute  tem- 
perature. The  absolute  temperature  is  273  degrees  plus  the  cen- 
tigrade temperature;  but  really  the  determination  of  the  absolute 
scale  of  temperature  depends  entirely  on  the  behavior  of  gases. 

The  original  statement  of  Charles'  Law  was  made  in  this  way: 
For  every  degree  rise  or  fall  in  temperature  the  volume  of  a  gas 
increases  or  decreases  by  an  amount  equal  to  -g-fg-  its  volume  at  0°  C. 
If  this  law  held  rigidly  all  the  way  down  the  scale  of  course  the 
volume  of  a  gas  would  become  zero  at  —  273°  C.  As  different 
gases  were  studied  it  was  found  that  they  obeyed  this  law  pretty 
exactly  until  they  got  near  the  temperature  at  which  they  would 
condense  to  a  liquid.  The  more  difficultly  condensible  a  gas,  the 
further  down  the  scale  it  would  follow  this  law,  and  finally  helium 


NOTES  AND  PROBLEMS  35 

which  was  the  last  gas  to  succumb  to  efforts  at  liquefaction,  fol- 
lows the  law  with  a  good  deal  of  accuracy  down  to  within  a  very 
few  degrees  of  —  273°  C.  Hence,  since  it  was  found  that  the 
less  condensible  a  gas  the  more  nearly  it  approximated  a  certain 
ideal  behavior,  a  so-called  "  perfect  gas  "  was  postulated  which 
would  have  exactly  the  ideal  behavior.  The  absolute  zero  then 
is  denned  as  the  temperature  at  which  the  volume  of  this  perfect 
gas  would  become  zero,  that  is  —  273°  C. 
Charles'  Law  is  expressed  in  the  equation 

v  =  T  X  k     (pressure  is  constant) 

where  T  stands  for  the  absolute  temperature,  where  k  is  a  constant 

value,  and  where  the  pressure  is  invariable.     If  the  volume  of 

the  gas  is  held  constant  then  it  is  the  pressure  which  must  vary 

p  =  T  X  k     (volume  is  constant) 

When  both  pressure  and  volume  may  vary  as  well  as  the  tem- 
perature we  can  use  the  equation 


_  p2v2 
Ti   '''"  T2 

which  embraces  both  Boyle's  and  Charles'  Laws. 

Dalton's  Law.  This  law  states  the  behavior  of  mixtures  of 
gases  as  follows:  When  two  or  more  gases  are  contained  in  the 
same  vessel  each  one  exerts  the  same  pressure  as  if  it  occupied 
the  whole  vessel  alone  at  that  temperature.  The  actual  meas- 
ured pressure  is  the  sum  of  the  partial  pressures  of  the  gases 
present. 

Since  the  gases  whose  volumes  we  ordinarily  have  to  measure 
are  confined  over  water  in  the  measuring  vessel,  and  are  conse- 
quently mixed  with  water  vapor,  we  are  especially  interested  in 
applying  Dalton's  law  to  water  vapor.  But  before  we  can  do 
this  we  must  know  the  amount  of  water  vapor  and  this  can  be 
determined  at  once  if  only  one  condition  is  met,  that  is  that  the 
gas  is  saturated  with  water  vapor. 

Saturated  Water  Vapor.  In  order  to  help  us  make  clear  the 
properties  of  saturated  water  vapor,  let  us  make  use  of  the  some- 
what idealized  apparatus  shown  in  Figure  13. 

The  piston  which  is  assumed  to  be  absolutely  gas  tight  is  raised 


36 


QUANTITATIVE  ASPECTS  OF  CHEMISTRY 


Piston    _LJ 


Cylinder  -»- 


Water 
m  Reservoir 


UJ 


Pressure 
^  Gauge 


Air  Pump 


to  a  middle  position  in  the  cylinder  and  fastened  there.  The 
stop  cock  to  the  vacuum  pump  is  opened  and  the  cylinder  evacu- 
ated until  the  pressure  gauge  reads  zero.  Now  the  cock  to  the 

pump  is  closed  and  the  cock 
from  the  water  reservoir  is 
opened  to  let  a  thin  layer  of 
water  run  into  the  bottom  of 
the  cylinder.  Immediately 
the  pressure  gauge  jumps  and 
shortly  adjusts  itself  to  exactly 
17.4  mm.  when  the  tempera- 
ture of  the  apparatus  is  20°  C. 
Now  let  us  raise  the  piston 
to  the  top  of  the  cylinder. 
There  is  a  momentary  depres- 
sion of  the  gauge  but  it  at  once 
returns  to  exactly  17.4  mm. 
Now  let  us  force  down  the 
piston  to  the  bottom  of  the 
cylinder.  Again  there  is  a 
slight  flicker  in  the  gauge  level 
but  the  reading  immediately 
becomes  exactly  17.4  mm. 
again,  if  the  temperature  is  maintained  constant  all  the  time 
at  20°. 

Saturated  water  vapor  is  in  equilibrium  with  liquid  water.  If 
the  vapor  is  not  saturated  and  any  liquid  water  is  present,  enough 
of  this  evaporates  to  make  the  vapor  saturated.  Compressing 
a  saturated  vapor  would  tend  to  increase  its  concentration  which 
would  increase  its  pressure.  The  observed  fact  that  there  is  no 
such  increase  in  pressure,  shows  that  the  vapor  does  not  become 
more  concentrated,  that  is,  that  it  does  not  become  supersaturated. 
Vapor  must  therefore  condense  to  liquid  water  to  maintain  the 
exact  state  of  saturation  as  the  piston  is  pressed  down;  and  when 
the  piston  is  drawn  up  the  liquid  must  evaporate  fast  enough 
to  maintain  a  saturated  condition  of  the  vapor. 

At  20°  the  concentration  of  saturated  water  vapor  is  such  that 
the  pressure  is  17.4  mm.  At  other  temperatures  the  concen- 
tration is  different  but  it  has  an  absolutely  definite  value  for  each 
temperature.  The  pressure  of  saturated  water  vapor  has  been 


FIG.  13 


NOTES  AND  PROBLEMS  37 

carefully  measured  at  all  temperatures  and  the  values  from  0° 
to  100°  are  printed  in  the  table  on  page  310  in  the  Appendix. 

We  shall  now  inquire  as  to  the  behavior  of  the  water  vapor  if 
the  cylinder  contains  also  some  permanent  gas,  for  example, 
oxygen.  Dalton's  law  would  tell  us  that  if  the  cylinder  contained 
the  same  amount  of  water  vapor  its  partial  pressure  would  be 
exactly  the  same  as  if  no  oxygen  were  present,  and  that  the  total 
pressure  of  the  gas  in  the  piston  would  be  the  sum  of  the  pressure 
of  the  oxygen  and  the  pressure  of  the  water  vapor. 

Experimentally  we  can  test  this  deduction.  Let  dry  oxygen 
into  the  dry  cylinder  until  the  gauge  stands  at  760  mm.  at  20°. 
Then  run  a  thin  layer  of  water  into  the  bottom  of  the  cylinder. 
The  gauge  begins  slowly  to  rise  and  finally  it  stops  at  777.4  mm. 
if  the  temperature  is  still  20°.  Thus  the  partial  pressure  of  sat- 
urated water  vapor  is  17.4  mm.  and  is  not  affected  by  the  pres- 
ence of  the  permanent  gas.  It  takes  a  much  longer  time  to  sat- 
urate the  space  in  the  cylinder  when  oxygen  is  present  because 
the  water  vapor  has  to  diffuse  through  the  oxygen,  but  the  final 
result  is  exactly  the  same.  The  oxygen  does  not  diminish  the 
capacity  of  the  space  for  the  water  vapor. 

Suppose  now  that  we  have  to  measure  a  quantity  of  oxygen  which 
has  been  collected  in  a  measuring  tube  over  water  in  a  trough. 
The  oxygen  has  bubbled  up  through  the  water  and  we  may  be 
certain  that  it  has  in  this  way  become  fully  saturated  with  water 
vapor.  We  raise  or  lower  the  measuring  tube  until  the  level  of 
the  water  is  the  same  inside  and  outside  the  tube.  On  the  outside 
surface  of  the  water  the  atmosphere  is  exerting  its  pressure  and 
this  pressure  is  transmitted  through  the  liquid  to  the  gas  within 
the  tube.  Let  us  say  that  the  barometer  reads  740  mm.  Then 
the  pressure  of  the  gas  is  740  mm.  But  this  pressure  is  the  total 
of  two  partial  pressures,  that  of  the  oxygen  and  that  of  the  water 
vapor.  Let  us  say  that  the  temperature  is  20°;  then  the  partial 
pressure  of  the  water  vapor  is  17.4  mm. ;  and  the  pressure  of  the 
oxygen  is  740  —  17.4  =  722.6  mm.  The  volume  read  in  the 
measuring  glass  is  say  60  cc.  Now,  remembering  Dalton's  law, 
we  can  say  that  we  have  an  amount  of  oxygen  which  at  20°  and 
722.6  mm.  occupies  a  volume  of  60  cc.  We  want  to  calculate  its 
volume  under  standard  conditions. 


760  ~  273 +20 


Vst    —    VobS    X   j_      i      r»TO    X 


38  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 

The  general  formula  for  reducing  to  standard  conditions  the  vol- 
ume of  a  gas  measured  over  water  is  therefore 

273      ^  p  —  aq.  tens. 
760 

Vat  =  volume  of  gas  under  standard  conditions. 
Voba  =  the  observed  volume  under  conditions  of  experiment, 
t  =  temperature  of  the  gas. 
p  =  pressure  upon  the  moist  gas. 
aq.tens.  =  tension  of  saturated  aqueous  vapor  at  t°  (see  table). 

PROBLEMS 

11.  Reduce  125.3  cc.  of  gas  at  725  mm.  to  760  mm.  pressure. 

12.  A  cylinder  of  2500  cc.  capacity  contains  oxygen  under 
91.5  atmospheres  pressure.     Find  volume  in  liters  after  the 
gas  is  run  into  a  tank  under  atmospheric  pressure. 

13.  Reduce  125.3  cc.  of  gas  at  25.8°  to  0°. 

.  14.  A  sealed  glass  tube  contains  125.3  cc.  of  gas  at  27° 
and  783  mm.  pressure.  What  will  be  the  pressure  if  the 
tube  is  heated  to  300°  without  change  of  volume? 

15.  Reduce  125.3  cc.  of  gas  at  740  mm.  and  20.7°  to  stand- 
ard conditions. 

16.  Reduce   125.3   cc.   of  gas  at   15.3   atmospheres  and 
—  65.5°  to  standard  conditions. 

17.  Reduce  125.3  cc.  of  gas  measured  over  water  at  740 
mm.  and  20.7°  to  standard  conditions. 

18.  A    certain   quantity   of   dry   hydrogen   gas   occupies 
2275  liters  at  25°  and  760  mm.     If  this  gas  were  bubbled 
through  water  and  collected  in  a  vessel  over  water,  what 
volume  would  it  then  occupy  at  the  same  temperature  and 
the  same  barometric  pressure?    Assume  that  no  hydrogen 
is  dissolved  in  the  water. 

Gay-Lussac's  Law  of  Combining  Volumes.  The  measurement 
of  the  volumes  of  gases  which  enter  into  chemical  reaction  led 
Gay-Lussac  to  perceive  the  existence  of  an  extremely  simple 
relationship  which  is  known  by  the  above  title  and  which  may  be 
stated  as  follows:  When  gaseous  substance  enter  into  chemical 
reaction  with  each  other  their  volumes  bear  the  relation  to  each 
other  of  small  whole  numbers;  furthermore,  if  the  products  of 


NOTES  AND  PROBLEMS  39 

the  reaction  are  also  gases  their  volumes  are  also  in  the  relation 
of  small  whole  numbers  to  each  other  and  to  the  volumes  of  the 
original  gases. 

Gay-Lussac  was  unable  to  explain  this  law  on  the  ground  of 
any  reasonable  hypothesis.  He  tried  to  postulate  that  equal 
volumes  of  different  gases  must  contain  the  same  number  of 
atoms,  but  this  postulate  was  almost  at  once  found  to  be  un- 
tenable and  he  had  nothing  better  to  offer. 

Avogadro's  Principle.  In  1811  Avogadro  offered  the  hypothesis 
that  at  the  same  temperature  and  pressure  all  gases  contain  in 
equal  volumes  an  equal  number  of  molecules.  He  defined  the 
molecule  as  the  smallest  particle  of  a  substance  and  drew  a  clear 
distinction  between  molecules  and  atoms;  a  molecule  of  a  com- 
pound would  of  necessity  contain  two  or  more  atoms,  at  least  one 
atom  of  each  constituent  element.  But  the  startling  feature  of 
Avogadro's  hypothesis  was  that  it  demanded  as  a  necessary  deduc- 
tion that  the  molecules  of  the  elementary  gases,  oxygen,  nitrogen, 
hydrogen,  and  chlorine  should  consist  of  two  atoms  each.  Such 
an  idea  was  regarded  by  his  contemporaries  as  preposterous  and 
they  would  have  nothing  to  do  with  his  hypothesis. 

It  was  therefore  not  until  1860  that  Avogadro's  hypothesis 
was  really  taken  seriously.  Since  then  it  has  become  one  of  the 
most  important  principles  of  Chemistry. 

Demonstration  that  Molecules  of  Elementary  Gases  are  Di- 
atomic. Let  us  consider  the  following  data  which  has  been  es- 
tablished by  measuring  combining  volumes  and  of  course  reducing 
all  volumes  to  standard  conditions: 

2  volumes  of  hydrogen  +  1  volume  of  oxygen  give  2  volumes 
of  water  vapor. 

1  volume  of  hydrogen  -f-  1  volume  of  chlorine  give  2  volumes 
of  hydrogen  chloride. 

2  volumes  of  ammonia  give  3  volumes  of  hydrogen  +  1 
volume  of  nitrogen. 

In  the  first  set  of  data  let  us  assume  that  the  one  volume  of 
contains  1  million  molecules.  Then  the  two  volumes 
of  water  vapor  according  to  Avogadro's  hypothesis  will  contain 
2  million  molecules  of  water.  But  the  water  vapor  is  a  homoge- 
neous substance  and,  since  it  contains  oxygen  at  all,  every  molecule 
of  it  must  contain  at  least  one  atom  of  oxygen.  Therefore  there 


40  QUANTITATIVE  ASPECTS  OF  CHEMISTRY1 

must  be  at  least  2  million  atoms  of  oxygen  which  were  derived 
from  the  1  million  molecules.  Therefore  every  molecule  of  oxy- 
gen must  contain  at  least  two  atoms. 

According  to  the  same  line  of  reasoning  1  million  molecules  of 
hydrogen  and  1  million  molecules  of  chlorine  produce  2  million 
molecules  of  hydrogen  chloride  which  must  contain  at  least  2 
million  atoms  of  hydrogen  and  2  million  atoms  of  chlorine.  There- 
fore the  molecules  of  hydrogen  and  of  chlorine  must  each  con- 
tain at  least  two  atoms. 

Since  two  volumes  of  ammonia  yield  only  one  volume  of  nitro- 
gen, the  same  reasoning  shows  that  the  molecule  of  nitrogen  must 
contain  at  least  two  atoms. 

Molecular  Weights.  Since  the  atomic  weight  of  oxygen  has 
been  settled  by  convention  as  16  and  since  there  are  at  least 
two  atoms  of  oxygen  in  a  molecule,  the  molecular  weight  of 
oxygen  must  be  at  least  32.  Let  us  neglect  the  qualification 
"  at  least "  and  take  the  molecular  weight  of  oxygen  as  32  with- 
out qualification,  making  this  our  standard  from  which  to  reckon 
all  other  molecular  and  atomic  weights. 

On  this  basis  the  molecular  weight  of  any  gaseous  substance  is 
at  once  determined  if  we  know  the  weight  of  a  measured  volume 
of  that  gas  under  any  stated  conditions.  Because  from  this,  the 
weight  of  1  liter  of  the  gas  under  standard  conditions  can  be  com- 
puted, and  the  weight  of  1  liter  of  the  gas  is  to  the  weight  of  1 
liter  of  oxygen  as  the  molecular  weight  of  the  gas  is  to  32,  the 
molecular  weight  of  oxygen. 

Gram  Molecular  Weight;  Mole.  The  molecular  weight  is 
an  abstract  number  although  the  molecular  weights  are  to  each 
other  in  the  same  ratio  as  the  actual  weights  of  the  molecules. 
The  gram  molecular  weight  is  however  a  concrete  quantity,  being 
the  molecular  weight  number  in  grams.  It  is  a  quantity  that  is  so 
frequently  spoken  of  that  the  name  has  been  contracted  to  mole. 
A  mole  of  oxygen  is  32  grams  of  oxygen. 

Gram  Molecular  Volume  or  Molal  Volume.  One  mole,  or 
32  grams  of  oxygen  under  standard  conditions  occupies  a  volume 
of  22.4  liters.  Since  according  to  Avogadro's  principle,  the  same 
volume  would  contain  exactly  1  mole  of  any  other  gas  under 
standard  conditions,  this  volume  assumes  a  great  importance  and 
it  is  named  the  gram  molecular  volume  or  for  brevity  molal  volume. 

To  find  the  molecular  weight  of  any  gaseous  substance  it  is 


NOTES  AND  PROBLEMS 


41 


necessary  therefore  only  to  find  the  weight  in  grams  of  22.4  liters 
of  the  gas  under  standard  conditions. 

The  actual  number  of  molecules  of  a  gas  in  22.4  liters  under 
standard  conditions  is  6.06  X  1023.  Although  it  has  been  only 
within  the  last  two  decades  that  this  number  has  been  ascertained, 
yet  it  is  named  Avogadro's  number  in  honor  of  the  man  who  reas- 
oned out  the  principle  on  which  this  number  rests. 

Avogadro's  number  is  the  number  of  molecules  in  a  mole  of  a 
substance :  it  is  also  the  number  of  atoms  in  a  gram  atomic  weight 
of  an  element. 

Atomic  Weights.  In  an  earlier  section  we  found  the  atomic 
weight  of  an  element  by  multiplying  16,  the  atomic  weight  of 
oxygen,  by  the  combining  ratio  of  the  element  with  oxygen  and 
dividing  by  the  assumed  atomic  ratio.  We  are  now  in  a  position 
to  verify  the  atomic  weights  so  found  and  thus  to  prove  whether 
the  assumed  atomic  ratio  was  correct. 

We  shall  start  with  the  molecular  weight  of  oxygen  taken  as  32 
and  first  demonstrate  that  on  this  basis  the  atomic  weight  of 
oxygen  is  16. 

There  are  hundreds  of  gaseous  compounds  of  oxygen  whose 
densities  have  been  measured  and  whose  molecular  weights  are 
thus  known.  The  molecules  of  these  compounds  can  contain 
a  small  whole  number  of  atoms  of  oxygen  but  never  fractional 
numbers.  The  percent  by  weight  of  oxygen  in  these  compounds 
can  be  found  by  chemical  analysis.  This  percent  of  the  molal 
weight  will  be  the  number  of  grams  of  oxygen  in  a  mole  of  the 
compound  and  this  number  must  be  either  the  atomic  weight  of 
oxygen  or  some  small  even  multiple  thereof.  In  the  following 
table  a  few  typical  examples  are  shown  from  a  collection  of  the 
data  for  all  gaseous  or  volatile  oxygen  compounds. 


Weight  in 

Percent  by 

Weight  of 

Greatest 

Substance 

grams  of  1 

weight  of 

oxygen  in  1 

common 

mole 

oxygen 

mole 

divisor 

Water              

18.016 

88.81 

16 

1  X  16 

Carbon  dioxide  .... 

44 

72.7 

32 

2  X  16 

Sulphur  trioxide  

80 

60 

48 

3  X  16 

Acetic  acid  

60 

53.3 

32 

2  X  16 

Phenol  

94 

17.02 

16 

1  X  16 

Nickel  carbonyl  

170.8 

37.5 

64 

4  X  16 

Phosphorus  oxy- 

chloride 

153.4 

10.43 

16 

1  X  16 

Oxygen               

32 

100. 

32 

2  X  16 

42 


QUANTITATIVE  ASPECTS  OF  CHEMISTRY 


The  smallest  number  in  the  next  to  the  last  column  is  16  and 
this  must  be  the  atomic  weight  of  oxygen  because  it  represents 
the  smallest  weight  of  oxygen  in  the  molecular  weight  of  any  of  its 
compounds.  It  is  extremely  unlikely  that  there  would  not  be 
at  least  some  compounds  whose  molecules  contained  but  a  single 
atom.  But  if  such  could  be;"  the  case  and  16  were  for  example 
twice  the  atomic  weight,  we  should  expect  that  8  would  be  the 
greatest  common  divisor  and  that  numbers  equal  to  3  X  8  or 
5X8  and  therefore  not  divisible  by  16  would  be  found  in  the 
next  to  the  last  column.  We  can  therefore  be  certain  that  16 
is  the  actual  atomic  weight  and  not  a  multiple  of  the  atomic  weight 
of  oxygen. 

A  similar  study  of  other  elements  for  example  of  chlorine,  hy- 
drogen, sulphur,  mercury,  shows  that  in  all  the  gaseous  or  volatile 
compounds  of  these  elements  the  smallest  weight  ever  found  in 
the  molal  volume  is  35.46  grams  of  chlorine,  1.008  grams  of  hydro- 
gen, 32.06  grams  of  sulphur,  and  200  grams  of  mercury;  further- 
more when  the  molal  volume  contains  a  greater  weight  of  these 
elements,  the  weight  is  in  every  case  a  small  even  multiple  of  these 
smallest  weights.  Thus  the  atomic  weights  of  these  elements 
as  given  in  the  table  are  verified. 

The  following  table  shows  selected  examples  from  the  col- 
lection of  the  data  for  all  gaseous  or  volatile  chlorine  compounds. 


Substance 

Weight  in 
grams  of  1 
mole 

Percent  by 
weight  of 
chlorine 

Weight  of 
chlorine  in  1 
mole 

Greatest 
common 
divisor 

Hydrogen  chloride  .... 
Chlorine.  . 

36.5 
70.9 

97.22 
100 

35.5 
70.9 

1  X  35.5 
2  X  35.5 

Mercuric  chloride  
Arsenic  trichloride.  .  .  . 
Silicon  tetrachloride  .  . 
Phosphorus  pentar 
chloride  

270.9 
182.1 
170.2 

208.3 

26.19 

57.7 
83.4 

85.2 

70.9 
107.0 
142 

177.5 

2  X  35.5 
3  X  35.5 
4  X  35.5 

5  X  35.5 

Derivation  of  a  Formula.  Let  us  take  for  example  acetic  acid. 
This  substance  is  a  liquid  at  ordinary  temperature  and  pressure 
but  it  is  easily  vaporized.  We  must  first  find  its  molecular  weight. 
By  measurement  it  is  found  that  1  liter  of  the  vapor  at  136.5°  C 


NOTES  AND  PROBLEMS  43 

and  760  mm.  weighs  1.786  grams.     Reduce  the  volume  to  stand- 

ard conditions: 

970 

=  1.000  X         *  ^36.5  °0.6667  liter. 


Thus  0.6667  liter  under  standard  conditions  weighs  1.786  grams. 
22.4  liters  will  weigh 

X  1.786  =  60.01  grams. 


Thus  the  molecular  weight  is  60.01. 

Chemical  analysis  shows  that  the  composition  of  acetic  acid  is: 
C,  40  percent;  H,  6.72  percent;   0,  53.3  percent. 
The  weight  of  each  of  these  elements  in  one  mole  is: 

C:  .40  X  60.01  =  24.00  =  2  X  12.01  grams 
H:  .0672  X  60.01  =  4.032  =  4  X  1.008  grams 
O:  .533  X  60.01  =  31.99  =  2  X  16.00  grams. 

Therefore  the  mole  of  acetic  acid  contains  2  gram  atomic  weights 
of  carbon,  4  gram  atomic  weights  of  hydrogen,  and  2  gram  atomic 
weights  of  oxygen  and  the  formula  is  C2H4O2. 

PROBLEMS 

19.  One  liter  of  a  certain  gas  under  standard  conditions 
weighs  2.25  grams.     Calculate  the  molecular  weight  of  this 
gas. 

20.  The  molecular  weight  of  a  certain  volatile  substance 
is  to  be  determined:   0.435  gram  of  the  substances  is  placed 
in  an  evacuated  vessel  and  the  whole  is  heated  to  136.5°,  at 
which  temperature  the  substance  is  entirely  converted  to  gas. 
The  pressure  and  volume  of  the  gas  are  now  found  to  be 
380  mm.  and  405.6  cc.,  respectively.     Calculate  the  molecu- 
lar weight  of  the  gas. 

21.  When  a  certain  liquid  substance  is  vaporized,  its  vapor 
is  found  to  weigh  5.413  times  as  much  as  an  equal  volume  of 
air  under  the  same  conditions.     Assuming  the  average  mo- 
lecular weight  of  air  to  be  28.955,  find  the  molecular  weight 
of  the  substance. 

22.  The  composition  of  the  above  substance  is  found  by 
analysis  to  be:   carbon,  45.87  percent;   hydrogen,  3.21  per- 


44  QUANTITATIVE  ASPECTS  OF  CHEMISTRY 

cent;   and  bromine,  50.92  percent.     Calculate  the  formula 
of  the  substance. 

23.  A  certain  compound  of  chlorine  and  copper  is  found  to 
have  the  composition:    copper,  64.2  percent;    chlorine,  35.8 
percent.     When  0.52  gram  of  this  substance   is  heated  to 
a  sufficiently  high  temperature  to  convert  it  completely  into 
a  gas,  it  takes  the  place  of  a  certain  volume  of  air.     This 
amount  of  air  is  found  to  measure  58.8  cc.  under  standard 
conditions.     Calculate  the  molecular  weight  and  the  chemi- 
cal formula  of  the  compound. 

24.  A    liquid    substance    has    the    composition:     carbon, 
12.76  percent;  hydrogen,  2.13  percent;  bromine,  85.11  per- 
cent;   and  when  vaporized  its  vapor  density  is  93.3  times 
that  of  hydrogen.     Calculate  the  molecular  weight  and  the 
formula  of  the  compound. 

25.  The  chloride  of  a  new  element  contains  38.11  percent 
of  chlorine  and  61.89  percent  of  the  element.     The  vapor 
density  of  the  compound  referred  to  air  is   12.85.     What 
is  the  atomic  weight  of  the  element  so  far  as  investigation  of 
this  one  substance  can  give  it? 

26.  Cyanogen   contains  46.08   percent  carbon  and  53.92 
percent  nitrogen.     Its  density  is  2.007   times  that  of  air. 
Calculate  its  formula. 

27.  0.5000  gram  of  a  substance  is  burned  in  a  current  of 
pure  oxygen  in  a  combustion  tube.     The  products  of  com- 
bustion are  passed  through  a  calcium  chloride  tube  which 
weighs  36.5011  and  36.7824,  respectively,  before  and  after 
the  experiment.     Find  the  percent  of  hydrogen  in  the  sub- 
stance. 

28.  The  products  of  combustion  in  No.  27  are  further 
passed  through  a  tube   containing  caustic  soda,   and  this 
weighs  40.4010  and  42.1184  grams,  respectively,  before  and 
after  the  experiment.     Find  the  percent  of  carbon  in  the 
substance. 

29.  The  above  substance  is  converted  into  a  vapor  at  273°, 
and  0.100  gram  is  found  to  occupy  a  volume  of  34.9  cc.  at 
760  mm.     Find  the  molecular  weight  of  the  substance. 

30.  From  the  results  calculated  in  Nos.  27,  28,  and  29 
find  the  formula  of  the  substance. 


CHAPTER      II 
WATER  AND   SOLUTION 

Water  is  in  many  respects  the  most  important  and  interesting 
substance  on  the  earth's  surface.  By  its  presence  in  abund- 
ance the  physical  and  chemical  conditions  necessary  for  the  ex- 
istence of  life  are  maintained  on  the  earth.  By  far  the  greater 
part  of  the  chemical  changes,  both  natural  and  artificial,  which 
have  a  direct  bearing  on  human  life  and  welfare  involve  water, 
either  as  a  direct  participant  in  the  change  or  as  a  solvent  for 
the  substances  which  are  changing. 

PREPARATION  1 

POTASSIUM  NITRATE  FROM  SODIUM  NITRATE  AND 
POTASSIUM  CHLORIDE 

Solubility  plays  a  controlling  part  in  many  chemical  processes 
of  which  the  present  one  is  a  typical  example. 

Salts,  as  well  as  strong  acids  and  strong  bases,  exist  in  a  pe- 
culiar state  when  they  are  in  solution.  This  is  known  as  the 
state  of  ionization  and  is  more  fully  dealt  with  in  Chapter  III, 
but  for  our  present  purposes  it  is  sufficient  to  know  that  salts 
consist  of  electrified  radicals,  the  metallic  radicals  being  positive 
and  the  non-metallic  radicals  negative.  These  radicals  are  held 
together  in  the  compound  by  electrostatic  attraction,  which  is 
enormous.  When  the  salt  is  in  solution,  the  electrostatic  at- 
traction still  exists  and  prevents  the  negative  radicals  from  get- 
ting away  from  positive  radicals,  but  there  is  a  certain  freedom 
of  movement  which  allows  radicals  to  exchange  places  easily, 
subject  only  to  the  condition  that  every  negative  radical  must  be 
electrically  balanced  at  all  times  by  some  positive  radical. 

When  sodium  nitrate  and  potassium  chloride  are  dissolved, 
the  solution  contains  four  ions  Na+,  N0s~,  K+,  Cl~,  and  from  these 
ions  not  only  could  the  two  original  salts  be  reconstructed,  but 
also  two  new  salts,  potassium  nitrate  and  sodium  chloride  through 
a  regrouping  of  the  radicals.  Which  of  the  four  salts  will  crys- 
tallize from  a  solution  containing  the  four  ions  depends  solely 

45 


46  WATER  AND  SOLUTION 

on  their  solubilities  and  the  latter  may  vary  much  or  little  with 
the  temperature.  The  following  table  and  diagram  give  the 

GRAMS  OF  SALT  SOLUBLE  IN  100  GRAMS  OF  WATER 

At  10°  At  100° 

KN03                           21  246 

NaCl                             36  40 

KC1                               31  56 

NaNO3                          81  180 

solubility  at  different  temperatures  for  each  salt.  For  example, 
at  10°,  21  grams  of  KNO3  are  soluble  in  100  grams  of  water.  This 
means  that  if  an  excess  of  solid  potassium  nitrate  is  shaken  with 
pure  water  until  no  more  will  dissolve  the  clear  solution  will  then 
contain  21  grams  of  KN03  for  every  100  grams  of  water.  On 
the  other  hand,  if  a  solution  of  42  grams  of  KN03  in  100  grams 
water  obtained  at  a  higher  temperature  is  cooled  to  10°  and  stirred 
for  a  while  to  bring  about  an  equilibrium  condition,  all  but  21 
grams  of  the  salt  crystallizes  out,  and  the  solution  has  exactly 
the  same  concentration  as  that  obtained  in  the  other  way. 

A  very  important  fact  concerning  solubilities  is  that  the  solu- 
bility of  a  given  salt  is  practically  unaffected  by  the  presence 
of  another  salt  in  the  solution  provided  only  that  the  other  salt 
does  not  possess  one  of  the  same  ions  as  the  first  salt. 

For  example,  suppose  sodium  nitrate  and  potassium  chloride 
in  equivalent  amounts  are  added  to  100  grams  of  water  so  that 
the  total  weight  of  K  and  NO3  radicals  will  be  42  grams,  the 
potassium  nitrate  in  excess  of  its  solubility  will  then  crystallize 
out  and  21  grams  of  the  crystals  will  thus  be  obtained.  The 
presence  of  the  radicals  of  sodium  chloride  in  the  solution  is 
without  effect  on  the  potassium  nitrate. 

In  the  following  procedure  equi-molal  amounts  of  sodium 
nitrate  and  potassium  chloride  are  taken  and  enough  water  added 
to  dissolve  at  the  boiling  temperature  all  of  the  sodium  nitrate 
taken  or  all  of  the  potassium  nitrate  which  could  result  from 
metathesis,  but  not  enough  to  dissolve  either  the  potassium 
chloride  taken  nor  the  sodium  chloride  which  could  be  formed 
by  metathesis.  Nevertheless  after  this  mixture  is  boiled  a  short 
time  all  of  the  solid  potassium  chloride  disappears  and  the  only 
solid  salt 'left  is  sodium  chloride. 


POTASSIUM   NITRATE  47 

The  mechanism  of  the  process  may  be  more  easily  appreciated 
if  it  is  represented  on  paper  in  the  following  fashion: 

KC1       ->  K+  Cl- 

NaNO3  ->  N03~        Na+ 

IT 

NaCl 

The  formulas  printed  in  bold  face  type  stand  for  the  sub- 
stances in  the  solid  state,  —  those  in  common  type  for  substances 
in  the  dissolved  state.  Single  arrows  indicate  that  the  reaction 
runs  to  completion  in  that  direction  under  the  conditions  pre- 
vailing (the  boiling  temperature  is  supposed  to  be  prevailing 
in  the  above  representation).  The  double  arrows  indicate  that 
an  equilibrium  is  reached  and  no  substance  shown  on  either  side 
of  the  arrows  disappears  from  the  sphere  of  action.  If  the  con- 
ditions were  to  be  shown  at  0°  solid  potassium  nitrate  would  have 
to  be  indicated  in  equilibrium  with  the  dissolved  salt. 

Materials:    crude  Chili  saltpeter  NaNO3,  170  grams  =  2  F.W. 

crude  potassium  chloride  KC1,  149  grams  =  2  F.W. 
Reagent:       1%  AgNOs  solution. 
Apparatus:  500-cc.  casserole. 

watch  glass. 

5-inch  funnel. 

perforated  filter  plate.  . 

800-cc.  suction  bottle  and  pump. 

platinum  wire. 

Procedure:  Place  the  sodium  nitrate  and  potassium  chloride 
in  a  500-cc.  casserole.  Add  210  cc.  of  water,  cover  with  a  watch 
glass,  and  place  over  a  low  flame.  Keeping  watch  of  the  casse- 
role to  see  that  the  contents  do  not  boil,  prepare  a  suction  filter 
according  to  Note  4  (b),  on  page  6.  Then  raise  the  flame  under 
the  casserole  and  watch  it  until  boiling  commences.  Lower  the 
flame  and  let  the  mixture  boil  gently  just  one  minute,  keeping 
the  watch  glass  over  the  casserole  to  prevent  too  much  evapora- 
tion of  water.  While  it  is  at  the  boiling  temperature,  pour  (see 
Figure  1,  page  5)  the  mixture  from  the  casserole  onto  the  suction 
filter  after  first  starting  a  gentle  suction.  Quickly  scrape  most 
of  the  damp  salt  onto  the  filter  and  suck  out  as  much  of  the  liquid 
as  possible.  Then  return  the  solid  salt,  which  is  mostly  NaCl, 


48  WATER  AND   SOLUTION 

to  the  casserole.  Pour  the  solution  into  a  beaker  and  cool  it  to  15° 
or  below  by  setting  it  in  a  pan  of  cold  water  or  better  ice  water. 
Stir  the  crystallizing  solution  frequently  so  as  to  obtain  a  loose 
crystal  meal  which  is  easier  to  handle  and  drain  on  the  filter  than 
the  larger  crystals  that  would  otherwise  be  obtained.  Separate  the 
crystals  of  KN03  from  the  coM  liquor  by  means  of  the  suction 
filter,  observing  last  sentence  of  Note  3  on  page  0,  and  pour  the 
liquor  into  the  casserole  containing  the  first  crop  of  NaCl  crys- 
tals. Bring  the  solution  to  boiling  point  and  boil  gently  three 
minutes  without  a  watch  glass  over  the  casserole,  thus  allowing 
some  of  the  water  to  escape  by  evaporation.  Then  filter  at  the 
boiling  temperature  exactly  as  in  the  first  instance.  Cool  the 
filtrate  and  collect  a  second  crop  of  KN03  crystals,  adding  them 
to  the  first  crop  and  pouring  the  liquor  into  a  flask  labelled 
"  Mother  Liquors."  Examine  the  two  kinds  of  crystals,  tasting 
them  and  using  a  microscope.  Draw  pictures  in  the  note  book 
of  the  crystals  as  seen  in  the  microscope.  Dissolve  about  0.1 
gram  of  the  product  in  2  cc.  of  water  and  test  for  chloride  by  adding 
1  drop  of  AgN03  solution.  Considerable  chloride  will  be  found 
and  the  product  must  be  purified  by  recrystallization.  Weigh 
the  crystals  roughly  while  they  are  still  moist,  add  one  half  their 
weight  of  hot  water,  and  warm  until  solution  is  complete.  Cool 
to  below  15°  and  separate  the  crystals  from  the  mother  liquor, 
adding  the  latter  to  the  reserve  flask.  Test  as  above  to  see  if 
this  crop  of  crystals  is  free  from  chloride.  If  not  repeat  the 
recrystallization  as  many  times  as  is  necessary  to  get  a  perfectly 
pure  product.  A  little  of  this  when  dissolved  should  give  no 
turbidity  with  silver  nitrate  solution,  and  when  held  in  the  flame 
on  a  platinum  wire  should  color  it  the  violet  color  characteristic 
of  potassium  with  none  of  the  yellow  sodium  color.  Spread  the 
preparation  on  paper  towels  and  allow  it  to  dry  by  standing 
exposed  to  the  air;  then  put  up  the  salt  in  a  2  ounce  bottle 
with  a  cork  stopper  and  label  it  neatly.  If  the  final  yield  of  pure 
product  is  not  satisfactory  in  amount  the  collected  mother  liquors 
should  be  boiled  down  to  about  100  cc.,  and  used  as  the  starting 
point  in  a  repetition  of  the  above  procedure.  50  grams  may  be 
regarded  as  a  very  satisfactory  yield. 

The  sequence  of  the  operations  in  this  preparation  can  be 
followed  rather  more  readily  in  the  tabulated  procedure  shown 
on  the  following  page. 


POTASSIUM  NITRATE 


49 


TABULATED  PROCEDURE 

Treat  170  grams  NaNO3  and  149  grams  KC1  with  210  cc.  water;  heat  to 
boiling  and  boil  one  minute;  filter  hot.  Do  not  rinse  out  dish  but  keep 
it  for  second  boiling. 


On  filter: 
NaCl,  dirt, 
some  KNO3: 
transfer  back 
to  dish  in 
which  first 
boiling  was 
made 
(1) 

Filtrate:    cool  and  filter 

Crystals: 
impure 
KNO3 

(2) 

Filtrate  is  saturated  with  KNO3  and  NaCl. 
Pour  into  dish  in  which  original  mixture 
was  boiled  and  to  which  impure  NaCl  (1) 
was  added.     Bring  to  boil,  boil  3  minutes, 
and  filter  hot. 

On  filter: 
NaCl 
and  dirt 
fairly  free 
from 
KNO3 
(3) 

Filtrate:  cool  and  filter 

Crystals: 
impure 
KNO3 

(4) 

Filtrate    is    satu- 
rated with  KNO3 
and  NaCl. 
Save  temporarily 
in  flask   labelled 
"Mother  Liquors" 
(5) 

RECRYSTALLIZATION 

Unite  impure  KNO3  (2)  and  (4) ;  heat  with  one  half  their  weight  of  water 
until  dissolved;  cool  without  filtering  and  collect  the  crystals  that  separate 
from  the  cold  solution. 


Crystals:    Nearly  pure  KNO3. 
Recrystallize    repeatedly    until    en- 
tirely  pure,    adding    all    mother    li- 
quors to  (5)  in  the  reserve  flask. 


Filtrate  contains  nearly  all  of  the 
NaCl  from  the  impure  product, 
and  is  saturated  with  KNO3;  add 
to  (5)  in  the  reserve  flask. 


Discard  mother  liquors  (5)  if  the  yield  of  pure  KNO3  is  satisfactory. 


QUESTIONS 

1.  Define  metathesis. 

2.  When  a  metathetical  reaction  is  carried  out  in  the  wet  way, 
why  is  the  solubility  of  the  substances  involved  of  importance? 
Explain  why,  according  to  this  point  of  view,  the  reactions  AgNO3 
+KC1  =  AgCl  +  KN03  and  BaCl2  +  Na^SO*  =  BaS04  +  2NaCl 
are  much  more  complete  than  the  reaction  NaN03  +  KC1  = 
KN03  +  NaCl. 

3.  Explain  why  fewer  operations  should  be  required  to  pre- 


50  WATER  AND   SOLUTION 

pare   potassium   nitrate   from   potassium   sulphate   and   barium 
nitrate  than  by  the  foregoing  procedure. 

4.  Explain  why  all  of  the  solid  salt  should  change  to  NaCl 
when  the  original  materials  are  boiled  with  insufficient  water  to 
dissolve  all  of  either  ihe  NaCLjar  the  KC1. 

5.  In  the  tabulated  procedure  what  is  the  advantage  of  adding 
the  impure  NaCl  (1)  to  the  second  mother  liquor  instead  of  dis- 
carding it?  .^ 

PREPARATION  2 

CRYSTALLIZED  SODIUM  CARBONATE  Na2CO3.10H2O  FROM 
ANHYDROUS  SODIUM  CARBONATE 

Many  substances,  as  acids,  bases,  salts,  oxides  and  even  ele- 
ments when  they  separate  from  solution,  or  crystallize,  carry 
water  with  them.  Such  substances  when  dried,  show  no  evi- 
dence of  fluiolity,  that  is  of  the  property  one  must  naturally 
think  of  as  belonging  to  water,  even  sometimes  when  more 
than  half  the  weight  of  the  substance  is  water.  Such  sub- 
stances are  called  hydrates  and  the  same  substance  without  the 
water  is  known  as  the  anhydrous  substance.  Substances  contain- 
ing water  of  this  nature  are  crystalline,  and  the  water  is  usually 
known  as  water  of  crystallization.  But  this  fact  should  not  lead 
one  to  think  that  water  is  necessary  for  the  formation  of  a  crystal, 
because  many  crystals  do  form  that  do  not  contain  water.  This 
water  is  held  in  the  hydrate  in  chemical  combination,  more  or 
less  firmly,  to  be  sure,  but  chemical  nevertheless,  because  such 
substances  always  obey  the  law  of  definite  proportions. 

Soda  ash  is  commercial  anhydrous  sodium  carbonate.  When  a 
solution  saturated  at  boiling  temperature  is  allowed  to  cool  to 
below  35°,  a  transparent  crystalline  mass  separates  of  the  formula 
Na2CO3.10H2O,  known  commercially  as  soda  crystals  or  sal  soda. 

This  substance  melts  at  about  35°  but  it  does  not  yield  a 
clear  liquid;  it  gives  rather  a  liquid  in  which  is  suspended 
a  solid  white  substance  of  the  formula  Na2CO3.H2O.  The  solu- 
tion obtained  by  melting  the  dekahydrate,  is  supersaturated  with 
respect  to  the  monohydrate,  and  the  latter  separates  out  leaving 
a  solution  which  contains  proportionately  more  water  than  the 
crystals  of  dekahydrate.  In  the  following  preparation,  if  barely 
enough  water  were  taken  to  form  the  dekahydrate/  a  complete 


CRYSTALLIZED  SODIUM  CARBONATE  51 

solution  could  not  be  obtained  because  some  monohydrate  would 
separate.  Enough  water  therefore  is  taken  to  hold  all  the 
monohydrate  in  solution.  Besides  the  monohydrate  and  the 
dekahydrate,  there  is  at  least  one  other  well  recognized  hydrate, 
the  heptahydrate  Na2C03.7  H2O.  This  might,  by  chance,  sepa- 
rate from  the  solution  in  the  following  procedure,  but  the  dekahy- 
drate is  far  more  likely  to  form.  If  after  the  solution  has  cooled 
to  below  35°,  the  melting  point  of  the  dekahydrate,  a  little 
crystal  fragment  known  to  consist  of  this  hydrate  is  added, 
this  will  serve  as  a  nucleus  for  the  crystallization  of  dekahydrate, 
and  if  any  heptahydrate  had  already  formed  it  would  soon  dissolve 
since  it  is  more  soluble  than  the  dekahydrate. 

Materials:      anhydrous  sodium  carbonate,  NaaCOs  1  mole  = 

106  grams. 
Apparatus:    500-cc.  flask. 

5-inch  filter  funnel. 

6-inch  crystallizing  dish. 

watch  glass  to  cover  dish. 

filter  plate. 

Procedure:  Place  the  sodium  carbonate  and  250  cc.  of  water 
in  the  flask  and  warm  it  to  just  short  of  the  boiling  point  until 
the  salt  is  dissolved.  See  Note  7,  p.  12.  Fold  an  8-inch  paper 
filter,  place  it  in  the  funnel,  and  pour  the  solution  through  the 
filter,  collecting  the  clear  filtrate  in  the  crystallizing  dish.  Cover 
the  latter  and  set  it  aside  to  cool  slowly.  After  six  hours  or 
longer  separate  the  crystals  from  the  remaining  liquid  by  pour- 
ing the  contents  of  the  dish  into  a  funnel  in  the  bottom  of  which 
is  placed  the  perforated  porcelain  filter  plate  without  any  paper. 
It  may  be  possible  to  get  additional  crystals  from  the  mother 
liquor  by  cooling  it  still  more.  After  the  crystals  have  drained 
transfer  them  to  some  paper  towels  and  dry  them  as  directed  in 
Note  9(6),  page  15.  When  they  are  dry  place  the  crystals  in  a 
6  ounce  cork-stoppered  bottle. 

QUESTIONS 

1.  What  reasons  have  you  for  believing  that  the  water 
in  crystal  hydrates  is  in  chemical  combination  with  the 
salt? 


52  WATER  AND  SOLUTION 

2.  How  could  anhydrous  sodium  carbonate  be  prepared  from 
the  hydrate? 

3.  How  could  the  monohydrate   Na2CO3.H2O   be   prepared? 
(See  preliminary  discussion  of  this  preparation.) 

PREPARATION  3 
POTASSIUM  AND  COPPER  SULPHATE.     K2S04.CuS04.6  H20. 

The  preceding  preparation  illustrated  the  formation  of  a  "  mo- 
lecular "  compound.  That  is,  starting  with  two  substances, 
water  and  sodium  carbonate,  in  which  the  ordinary  chemical 
combining  powers  of  the  elements  were  seemingly  satisfied,  it 
was  found  that  the  compounds  themselves  would  combine  with 
each  other.  By  this  combination  the  character  of  the  individual 
compounds  was  not  destroyed  altogether,  their  separate  mole- 
cules apparently  retaining  their  identity  even  in  the  combination. 
Hence  the  name  molecular  compound.  The  present  preparation 
shows  that  two  salts  can  form  a  molecular  compound  with  each 
other  and  with  water.  Potassium  sulphate  ordinarily  crystallizes 
from  solution  as  anhydrous  K2S04,  copper  sulphate  as  the  hy- 
drate CuSO4.5  H2O,  but  when  both  salts  are  together  in  solution 
a  compound  hydrate  separates  K2S04.CuSO4.6  H2O. 

Materials:      potassium  sulphate  35  grams  =  |  mole. 

blue  vitriol,  CuS04.5  H2O  50  grams  =  ^  mole. 
Apparatus:    8-inch  crystallizing  dish. 

5-inch  filter  funnel  and  filter  paper. 

thermometer. 

Procedure:  Grind  the  two  salts  together  in  a  mortar  until 
they  are  very  finely  powdered.  Add  the  mixture  to  250  cc.  of 
water  in  a  porcelain  dish;  raise  the  temperature  to  just  5°  above 
the  room  temperature  and  keep  it  at  that  point  during  10  min- 
utes, while  stirring  all  of  the  time  with  the  stem  of  a  thermometer. 
This  gives  a  practically  saturated  solution  of  the  double  salt. 
Let  the  undissolved  salt  settle  a  moment  and  pour  off  the  solution 
into  another  dish.  With  a  little  more  water  prepare  in  the  same 
way  a  saturated  solution  of  any  residue  that  is  left.  Add  a  few 
drops  of  sulphuric  acid  to  the  entire  solution,  warm  it  to  about 
50°,  and  filter  it,  allowing  the  filtrate  to  run  directly  into  the 
crystallizing  dish.  Set  the  solution  away  uncovered  to  crystallize 


HYDRATES  53 

in  a  place  protected  from  the  dust;  and,  before  it  has  cooled  to 
below  the  saturation  temperature,  seed  it  with  eight  or  ten  small 
crystals  of  the  double  salt.  Allow  the  solution  to  stand  until  a 
large  crop  of  crystals  has  formed.  Drain  the  crystals,  wrap 
them  in  paper  towels  and  dry  them  as  directed  in  Note  9(6), 
page  15.  If  only  a  small  amount  of  mother  liquor  is  left,  it  may 
be  discarded. 

It  is  a  matter  of  considerable  difficulty  to  obtain  clear,  well- 
formed  crystals  of  this  salt,  especially  if  the  laboratory  tempera- 
ture varies  a  good  deal.  If  the  conditions  are  too  unsatisfactory 
it  is  better  to  give  up  the  attempt  to  get  distinct  crystals  and  to 
prepare,  instead,  a  crystalline  meal:  Dissolve  the  same  amounts 
of  the  two  simple  salts  in  200  cc.  of  hot  water;  add  a  few  drops 
of  sulphuric  acid;  filter,  and  cool  the  solution  rapidly  in  a  flask 
while  rotating  it  under  the  water  tap.  Drain  the  crystal  meal 
on  a  suction  filter  and  dry  it  as  directed  above. 

QUESTIONS 

1.  Compare  the  color  of  crystals  of  the  double  salt  with  that  of 
the  simple  salts.     Likewise  compare  the  color  of  the  solutions. 
Does  this  indicate  that  the  double  salt  exists  in  solution? 

2.  Try  by  experiment  on  a  small  scale  whether  a  similar  double 
salt  containing  ammonium  sulphate  in  place  of  potassium  sulphate 
can  be  prepared.     Dissolve  1  gram  of  the  ammonium  sulphate 
and  2  grams  of  blue  vitriol  in  10  cc.  of  hot  water.     Pour  through 
a  filter  and  collect  the  solution  on  a  watch  glass.     When  crystals 
have  formed  see  if  they  are  all  of  the  same  kind  (i.e.  the  double 
salt)  or  of  two  distinct  kinds  (the  two  simple  salts).     Probably 
the  crystals  will  be  small  and  a  microscope  will  help.     Notice 
whether  in  color  and  shape  they  resemble  the  simple  salts  or  the 
double  salt  of  the  main  preparation. 

Repeat  the  same  experiment  with  sodium  sulphate  (1  gram 
lydrous  Na^SOJ  and  copper  sulphate. 

Experiments 
HYDRATES 

1.  Water  of  Crystallization.  Heat  a  crystal  of  blue  vit- 
riol CuSO4.5  H2O  rather  cautiously  in  a  dry  test  tube  hold- 
ing the  latter  in  a  nearly  horizontal  position.  Observe  that 


54  WATER  AND  SOLUTION 

the  crystal  gradually  loses  its  blue  color  and  becomes  white 
and  powdery.  Also  that  drops  of  water  condense  on  the 
cooler  part  of  the  wall  of  the  tube.  Let  the  tube  cool  to  room 
temperature  and  add  a  few  drops  of  water  to  the  white  cop- 
per sulphate.  Observe  t&at  the  material  regains  at  once 
its  original  blue  color  and  that  it  grows  so  hot  that  the  hand 
cannot  be  held  on  that  end  of  the  tube. 

Water  combined  as  water  of  crystallization  in  crystals  is  in  a 
true  state  of  chemical  combination,  for  the  hydrated  crystal  shows 
all  the  characteristics  of  a  chemical  compound. 

First:  its  physical  properties  such  as  color  (in  this  instance) 
and  crystalline  form  are  altogether  different  from  those  of  the 
anhydrous  salt  or  of  water. 

Second:  it  follows  the  law  of  definite  proportions,  every  sample 
of  blue  vitriol  for  example  having  exactly  the  composition  ex- 
pressed by  the  formula  CuS04.5  H2O.  (The  next  experiment  is  a 
quantitative  one  designed  to  prove  that  a  crystal  hydrate  has  a 
definite  composition.) 

Third:  there  is  a  marked  heat  effect  produced  by  the  combina- 
tion of  the  anhydrous  salt  and  water. 

2.  Composition  of  a  Crystal  Hydrate.  The  mineral  gyp- 
sum contains,  besides  calcium  sulphate,  also  a  certain  pro- 
portion of  water.  The  latter  may  be  completely  driven  off 
by  heat,  leaving  anhydrous  calcium  sulphate. 

Weigh  accurately  a  clean,  dry,  15  cc.  porcelain  crucible. 
Place  approximately  2  grams  of  gypsum  in  it  and  weigh  again. 
Cover  the  crucible,  support  it  on  a  clay  triangle,  and  heat 
(to  avoid  breaking  the  cover  warm  this  first  uniformly  by 
playing  the  flame  over  it  carefully  from  above)  to  redness 
for  20  minutes  in  the  Bunsen  flame.  Rest  the  triangle  and 
crucible  on  top  of  a  good-sized  beaker  and  allow  to  cool. 
Again  weigh  the  crucible  and  contents.  Repeat  the  heat- 
ing, and  if  a  further  loss  of  weight  occurs  repeat  until  two 
successive  weighings  are  the  same.  This  is  to  ensure  that  all 
the  water  is  driven  off.  The  cover  is  used  for  a  double  pur- 
pose: to  prevent  fragments  of  the  crystals  snapping  out  of 
the  crucible  from  sudden  expansion  of  the  steam,  and  to 
keep  the  heat  in. 

Calculate  the  percentage  by  weight  of  water  in  gypsum. 


HYDRATES  55 

From  this  result  calculate  the  number  of  molecules  of  water 
of  crystallization  in  gypsum,  assuming  the  formula  to  be 
CaS04.nH2O  (atomic  weights  in  appendix). 

If  the  experimental  work  is  carefully  done  it  will  be  found 
that  n  is  an  even  whole  number.  Thus  the  substances  calcium 
sulphate  and  water  combine  chemically  in  amounts  proportional 
to  simple  integral  multiples  of  their  molecular  weights  just  as 
elements  combine  in  amounts  proportional  to  simple  integral 
multiples  of  their  atomic  weights.  Although  the  crystal  hydrate 
is  just  as  truly  a  chemical  compound  as  is  calcium  sulphate  or 
water,  nevertheless  the  force  holding  the  substances  together  in 
this  compound  is  much  inferior  to  that  holding  the  elements 
together  in  the  simpler  compounds. 

3.  Efflorescence.  Leave  some  crystallized  sodium  sui- 
phate  Na2SO4.10H20  or  zinc  sulphate  ZnSO4.7  H2O  or  soda 
crystals  NasCOs.lO  H2O  uncovered  on  a  watch  glass  for  some 
time. 

Observe  that  the  surface  soon  becomes  white  and  powdery 
and  finally  the  whole  crystal  changes  to  a  white  powder. 

This  change  is  caused  by  the  loss  of  water  of  crystallization 
from  the  crystal,  leaving  either  the  anhydrous  salt  or  a  less  hy- 
drated  salt  (of  definite  composition  however).  If  the  experiment 
is  repeated  leaving  the  hydrated  salt  under  a  bell  jar  in  which  was 
also  set  a  dish  of  water  no  efflorescence  is  observed.  Pure  water 
exerts  a  higher  pressure  of  aqueous  vapor  than  does  the  crystal 
hydrate,  and  hence  the  hydrate  is  prevented  from  efflorescing. 
In  the  average  laboratory  air,  however,  the  pressure  of  aqueous 
vapor  is  rarely  over  50  percent  of  the  saturation  value.  If  the 
existing  pressure  is  less  than  that  which  the  hydrate  can  exert, 
the  hydrate  will  effloresce,  otherwise  not.  Blue  vitriol  exerts  a 
rather  low  vapor  pressure,  one  which  is  ordinarily  exceeded  by 
the  aqueous  vapor  in  the  air;  but  on  very  dry  days  the  pressure 
of  aqueous  vapor  in  the  air  sometimes  falls  below  that  of  the  blue 
vitriol  and,  on  such  days  only,  may  this  salt  be  observed  to  efflor- 
esce. 

Heating  blue  vitriol  increases  its  aqueous  tension  and  thus 
the  dehydration  by  heating  is  in  fact  an  artificially  induced  efflor- 
escence. 


56  WATER  AND  SOLUTION 

4.  Deliquescence.  Leave  small  lumps  of  calcium  chlo- 
ride, zinc  chloride,  potassium  hydroxide,  sodium  hydroxide, 
or  ferric  chloride  in  a  watch  glass  and  note  that  in  a  few 
minutes  the  surface  becomes  covered  with  liquid,  and  that 
after  a  longer  time  the  lumps  have  completely  changed  to 
liquid. 

This  liquefaction  is  due  to  condensation  of  the  water  vapor 
of  the  air  and  the  dissolving  of  the  solid  material  in  the  condensed 
water.  All  deliquescent  substances  are  extremely  soluble  in 
water.  The  tension  of  the  vapor  escaping  from  water  is  always 
lessened  by  substances  dissolved  in  the  liquid  water,  and  when 
this  tension  is  lowered  to  below  that  of  the  water  vapor  existing 
in  the  atmosphere  the  water  vapor  ceases  to  escape.  Indeed  the 
opposite  takes  place,  namely  water  vapor  condenses.  In  general 
substances  will  deliquesce  when  the  aqueous  tension  of  their 
saturated  solutions  is  less  than  the  tension  of  water  vapor  in  the 
air. 

If  these  deliquescent  substances  were  put  under  a  bell  jar, 
together  with  a  shallow  dish  containing  concentrated  sulphuric 
acid  to  absorb  the  water  vapor,  it  is  obvious  that  they  would  not 
deliquesce;  in  fact,  if  they  have  already  deliquesced  and  are 
placed  in  such  a  dry  atmosphere  the  water  will  evaporate  and 
they  will  again  change  to  dry  solid  substances. 

ELEMENTS  AND  WATER 

When  substances  like  salts  and  water  form  compounds  the 
force  of  attraction  is  of  a  subordinate  character,  and  the  formula 
of  a  hydrate  is  usually  written  with  a  period  between  the  formula 
of  the  salt  and  that  of  the  water.  This  indicates  that  the  "  pri- 
mary "  valence  of  the  element  is  satisfied  in  the  simple  compounds, 
and  that  it  is  only  some  sort  of  a  minor  or  "  secondary  "  valence 
which  holds  the  substances  in  combination. 

When  elements  react  with  water,  however,  a  change  in  the 
primary  valences  is  usually  involved. 

A  rather  unusual  instance  of  a  hydrate  of  an  element  is  chlorine 
hydrate,  C12.8  H2O,  which  crystallizes  from  ice  cold  water  which 
is  saturated  with  chlorine.  This  substance  is  clearly  a  molecular 
compound,  for  if  these  crystals  are  placed  in  a  watch  glass  and 
allowed  to  come  to  room  temperature,  chlorine  gas  escapes  and 


ELEMENTS  AND  WATER  57 

water  containing  only  the  amount  of  chlorine  corresponding  to 
an  ordinary  saturated  solution  is  left. 

5.  Sodium  and  Water.    Danger.    Fill  a  small  beaker  to 
about  2  cm.  depth  with   cold  water;    place   it  on  a  desk 
top    well    removed    from    any    bystander,    and    at    arms 
length  drop  a  small  bit  of  metallic  sodium  on  to  the  water. 
Note  that  the  sodium  floats;    that  it  almost  immediately 
melts  to  a  globule  with  a  bright  metallic  surface;   that  a 
gas  is  given  off  freely  under  the  impulse  of  which  the  glob- 
ule races  about  over  the  surface  of  the  water;    that  finally 
the  globule  entirely  disappears,  and  then  that  the  remaining 
solution  is  strongly  alkaline,  turning  litmus  blue  and  making 
the  fingers  slippery  when  wet  with  it.     Often  after  the  so- 
dium has  melted  and  become  hot  there  is  a  violent  explo- 
sion which  throws  the  caustic  solution  and  the  burning  metal 
about,  hence  the  danger.     This  explosion  is  accidental  and 
may  be  due  to  an  impurity  in  the  sodium. 

The  gas  evolved  is  hydrogen;  the  alkaline  character  of  the 
solution  is  due  to  sodium  hydroxide  which  is  left  dissolved  in  the 
water 

2  Na  +  2  HOH  -»  2  NaOH  +  H2 

Sodium  is  a  much  more  active  element  than  hydrogen  and  has 
displaced  an  equivalent  amount  of  the  latter  from  water.  So- 
dium hydroxide  then  may  be  regarded  as  water  in  which  sodium 
has  taken  the  place  of  one  half  of  the  hydrogen. 

6.  Calcium  and  Water.    Calcium  is  a  very  tough  metal 
and  is  difficult  to  cut  or  break.     A  lump  of  it  may  be  taken 
to  a  silversmith  and  rolled  into  a  thin  sheet  and  the  latter 
may  be  cut  into  small  pieces  with  tinsmith's  shears.     Drop 
one  or  two  small  bits  of  calcium  into  a  test  tube  of  clear  water. 
Note  that  a  gas  is  evolved;    collect  a  little  of  it  and  note 
that  it  burns  like  hydrogen.     Note  that  the  water  remains 
clear  for  a  little  time  but  that  soon  it  becomes  cloudy  due  to 
the  separation  of  a  white  finely  divided  solid  substance.    The 
solution  colors  litmus  blue,  but  produces  hardly  a  noticeable 
slippery  feeling  between  the  fingers. 


58  WATER  AND  SOLUTION 

This  experiment  is  similar  to  the  preceding  one.  Calcium  dis- 
places hydrogen  from  water 

Ca  +  2  HOH  -»  Ca(OH)2  +  H2 

but  the  reaction  is  less  violent*  calcium  is  thus  shown  to  be  a  less 
active  metal  than  sodium.  Furthermore  the  product  obtained 
by  substituting  calcium  for  hydrogen  in  water,  calcium  hydroxide, 
is  sparingly  soluble  and  appears  as  a  solid  -substance  as  soon  as 
more  than  enough  of  it  has  been  formed  to  saturate  the  solution. 
The  formula  Ca(OH)2  shows  that  an  atomic  weight  in  grams  of 
calcium  can  take  the  place  of  two  gram  atomic  weights  of  hydro- 
gen, in  other  words  the  valence  of  calcium  is  two. 

Although  it  is  only  the  most  active  metals  that  displace  hydro- 
gen freely  from  cold  water,  nevertheless  many  of  the  metals  do 
react  with  water  but  for  one  reason  or  another  the  reaction  does 
not  progress  far.  We  just  saw  that  calcium  hydroxide  was  only 
sparingly  soluble.  The  hydroxides  of  magnesium,  zinc,  aluminum, 
lead,  iron  are  even  less  soluble.  Thus,  although  a  freshly  cleaned 
piece  of  metal  may  react  with  water,  the  hydroxide  which  is 
produced  adheres  to  the  surface  as  a  coating  which  separates  the 
metal  and  the  water.  This  is  the  main  reason  why  most  of  the 
fairly  active  metals  seem  to  be  without  action  on  water. 

The  following  few  experiments  illustrate  the  action  of  such 
metals  with  water.  In  every  case  it  is  a  question  of  removing  or 
breaking  through  the  film  on  the  surface. 

7.  Magnesium  and  Water,  (a)  Note  the  appearance  of 
the  surface  of  some  magnesium  ribbon.  Scrape  the  surface 
with  a  knife  and  note  the  bright  metallic  luster.  Note  also 
that  this  luster  quickly  grows  dim.  Note  that  the  mag- 
nesium is  without  perceptible  action  on  water,  either  cold 
or  boiling. 

(6)  To  try  the  effect  of  hot  magnesium  in  dry  steam, 
clamp  a  piece  of  hard  glass  tubing  1  cm.  in  diameter  (ar- 
senic tubing)  in  position  at  an  angle  of  about  5°  with  the 
horizontal.  Connect  the  lower  end  of  the  tube  with  the 
wash  bottle  in  such  a  manner  that  steam  generated  by  boil- 
ing the  water  may  be  conducted  through  the  tube.  Connect 
the  upper  end  of  the  tube  with  a  delivery  tube,  so  that  the 
gases  produced  in  the  reaction  may  be  collected  over  water 


ELEMENTS  AND  WATER  59 

in  a  trough.  Place  some  pieces  of  magnesium  ribbon  in  the 
middle  of  the  hard  glass  tube.  Pass  steam  through  it,  and 
cautiously  commence  heating  at  the  lower  end  until  the 
condensed  water  is  evaporated  and  the  steam  is  "  dry." 
Get  the  whole  length  of  the  tube  well  warmed  with  the 
burner,  so  that  it  appears  perfectly  dry  inside  and  then  heat 
the  section  containing  the  magnesium  as  strongly  as  possible. 
It  is  rather  difficult  to  get  the  magnesium  to  catch  fire,  and 
it  may  be  necessary  to  stop  the  water  boiling  in  the  flask  for 
a  moment,  because  the  steam  is  so  much  colder  than  the 
kindling  temperature.  As  soon  as  the  magnesium  catches 
fire  make  the  steam  pass  vigorously  again  and  slip  a  test 
tube  full  of  water  over  the  mouth  of  the  delivery  tube.  No- 
tice that  the  magnesium  appears  to  burn  in  the  dry  steam 
with  much  the  same  brilliancy  as  in  air;  that  a  white  smoke 
and  ash  (magnesium  oxide)  is  produced  as  in  air.  The 
blackening  of  the  parts  of  the  glass  that  were  in  contact 
with  the  melted  magnesium  or  its  vapor  may  be  disregarded 
as  far  as  the  purpose  of  this  experiment  goes;  it  is  due  to 
a  reduction  of  the  silicon  dioxide  of  the  glass  to  silicon. 
The  gas  caught  in  the  trough  is  found  to  burn  with  a  color- 
less flame  and  is  thus  shown  to  be  hydrogen. 

(c)  Magnesium  amalgam  is  an  alloy  (much  like  a  solu- 
tion) of  mercury  and  magnesium.  It  is  made  by  rubbing 
powdered  magnesium  and  mercury  together  in  a  mortar. 
Since  a  good  deal  of  rubbing  is  necessary,  it  is  obvious  that 
the  oxide  film  on  the  surface  of  the  metal  has  to  be  rubbed  off 
before  the  mercury  can  begin  to  dissolve  the  metal.  When 
the  amalgamation  once  begins,  a  good  deal  of  heat  is  pro- 
duced and  the  process  is  soon  finished.  A  semi  liquid  or  a 
stiff  amalgam  is  obtained,  according  to  the  proportions. 

Drop  a  small  lump  of  magnesium  amalgam  into  a  test 
tube  of  cold  water.  A  violent  reaction  takes  place.  A  gas 
is  freely  evolved  which  burns  like  hydrogen.  The  con- 
tents of  this  tube  grow  very  hot  and  the  amalgam  disinte- 
grates to  yield  a  finely  divided  gray  powder  which  stays  sus- 
pended in  the  water. 

In  experiment   (b)  the  action  was  extremely  slow  until  the 
magnesium  reached  its  melting  point.     From  then  on  no  coherent 


60  WATER  AND  SOLUTION 

film  would  stick  to  the  liquid  surface,  and  without  this  mechanical 
hindrance  to  the  reacting  substances  coming  in  contact  with  each 
other,  the  natural  activity  of  the  magnesium  came  into  play. 

In  experiment  (c)  the  surface  of  the  amalgam  is  liquid  or  at 
least  semi  liquid  so  that  no  coherent  film  of  magnesium  hydroxide 
can  adhere. 

That  magnesium  must  be  a  vastly  more  active  element  than  hy- 
drogen is  shown  by  the  vigor  with  which  it  displaces  it;  in  ex- 
periment (b)  for  example  the  fact  that  the  oxygen  with  which  the 
magnesium  was  combining  had  to  be  withdrawn  from  its  com- 
bination with  hydrogen  did  not  appear  to  sensibly  diminish  the 
vigor  of  the  reaction  for  the  incandescence  was  practically  as 
bright  as  if  the  magnesium  ribbon  had  been  burning  in  oxygen. 

8.  Iron    and    Water.     Iron    filings    do   not   produce   any 
measurable  evolution  of  hydrogen  either  in  cold  or  hot  water. 
Arrange  to  pass  steam  through  a  tube  arranged  as  in  ex- 
periment 76  and  containing  iron  filings.     Heat  cautiously 
to  dry  the  liquid  water  and  then  heat  the  iron  filings  strongly. 
As  the  steam  from  the  delivery  tube  condenses  in  the  water 
in  the  trough  it  is  now  seen  that  an  occasional  bubble  of  gas 
rises.     Before  long  enough  collects  to  test  and  it  proves  to 
burn  like  hydrogen. 

Apparently  in  this  case  the  coating  of  oxide  on  the  surface 
of  the  iron  is  not  entirely  impervious  to  gases.  At  the  high 
temperature  the  steam  diffuses  more  rapidly  through  the  layer 
and  thus  the  production  of  hydrogen  becomes  measureably  rapid. 

9.  Removal   of   protective    coating   by    chemical   action. 

Note  that  aluminum  metal  is  like  magnesium  in  that  it 
displays  a  brilliant  metallic  luster  when  its  surface  has  just 
been  scraped  but  it  quickly  loses  the  brightness  of  its  luster. 
Drop  some  aluminum  turnings  into  some  sodium  hydroxide 
solution.  Note  that  gas  is  immediately  given  off,  which  a 
test  shows  to  burn  like  hydrogen  and  that  the  aluminum  in 
time  dissolves  completely. 

Aluminum  hydroxide  is  known  to  react  readily  with  sodium 
hydroxide  producing  the  soluble  sodium  aluminate 

A1(OH)8  +  3  NaOH  ->  Na3AlO3  +  3  H20. 


OXIDES  AND  WATER  61 

With  the  aluminum  hydroxide  coating  thus  continually  removed 
in  this  way  there  is  nothing  to  prevent  the  progress  of  the  primary 
reaction 

2  Al  +  6  H20  ->  2  A1(OH)8  +  3  H2. 

10.  Chlorine  and  Water.     The  chlorine  water  in  the  re- 
agent bottle  is  prepared  by  dissolving  chlorine  gas  in  water. 
Take  a  few  cubic  centimeters  of  this  solution  and  (to  make 
sure  that  it  has  been  acted  upon  by  light)  expose  it  a  few  min- 
utes to  direct  sunlight  or  for  a  longer  time  to  strong  diffused 
daylight.     Boil  the  solution  under  the  hood  to  drive  off  any 
chlorine  which  is  left,  and  then  test  with  litmus.     The  litmus 
is  turned  red.     Fill  a  test  tube  with  chlorine  water  and  in- 
vert it  in  a  small  beaker  containing  chlorine  water.     Leave 
the  whole  in  the  bright  sunlight  for  several  hours.     A  few 
cc.  of  a  colorless  gas  collects  in  the  top  of  the  inverted  test 
tube,  and  the  yellow  color  of  the  chlorine  water  gradually 
fades  out.     The  gas  causes  a  glowing  splinter  to  burst  into 
flame  and  is  thus  shown  to  be  oxygen. 

Metals  more  active  than  hydrogen  displaced  that  element  from 
water.  Non-metals  on  the  other  hand  if  active  enough  would 
displace  the  oxygen.  It  is  obvious  that  this  is  what  has  happened 
and  that  the  acid  remaining  in  solution  is  hydrochloric  acid,  HC1. 

The  total  effect  of  the  change  is  given  by  the  equation 

2  C12  +  2  H2O  -»  4  HC1  +  O2. 

That  chlorine  is  not  greatly  more  active  than  oxygen  is  shown  by 
the  fact  that  this  reaction  does  not  take  place  in  the  dark,  but 
only  under  the  action  of  sunlight. 

OXIDES  AND  WATER 

Nearly  all  of  the  elements  are  capable  of  combining  with  oxygen 
to  form  oxides,  but  there  seems  to  be  some  sort  of  a  residual  com- 
bining power,  for  all  of  the  oxides  have  a  greater  or  less  tendency 
to  combine  with  water  or  with  other  oxides.  The  oxides  of 
metals  combine  with  water  to  form  bases:  the  oxides  of  non- 
metals  combine  with  water  to  form  acids. 

11.  Sodium  Oxide  and  Water.     Support  a  porcelain  cru- 
cible cover  or  a  broken  piece  of  a  porcelain  dish  on  a  triangle 


62  WATER  AND  SOLUTION 

and  heat  it  to  redness;  while  it  is  thus  hot,  place  upon  it, 
by  means  of  iron  pincers,  a  small  piece  of  sodium  the  size 
of  a  small  pea.  Remove  the  lamp  and  let  the  sodium 
burn.  When  cold  dissolve  the  oxide  in  a  few  cubic  centi- 
meters of  water  and  test  the  solution  with  litmus.  Notice 
some  effervescence  when  the  oxide  is  dissolving.  Note  that 
the  solution  turns  litmus  blue  and  feels  very  slippery  when 
rubbed  between  the  fingers. 

When  sodium  burns  in  an  abundance  of  air  an  oxide  of  the 
formula  Na202  and  called  sodium  peroxide  is  formed.  This 
oxide  has  twice  as  much  oxygen  to  a  given  weight  of  sodium  as 
another  oxide  Na2O.  The  latter  can  be  prepared  by  heating 
together  sodium  peroxide  and  sodium  out  of  contact  with  the  air. 
When  sodium  peroxide  is  treated  with  water  it  reacts  according 
to  the  equation 

2  Na*02  +  2  H2O  ->  4  NaOH  +  02 

thus  losing  one  half  of  its  oxygen  and  yielding  the  same  soluble 
product  as  when  the  ordinary  oxide  of  sodium  combines  with 
water 

Na20  +  H20  ->  2  NaOH. 

Sodium  hydroxide,  NaOH,  is  very  soluble  in  water;  it  is  one  of  the 
strongest  bases. 

12.  Calcium  Oxide  and  Water.  The  well  known  sub- 
stance quick  lime  is  calcium  oxide.  Calcium  oxide  could  be 
made  by  burning  bits  of  the  metal  calcium,  but  the  oxide  so 
coats  over  the  surface  of  the  lump  that  it  is  difficult  to  make 
the  interior  portions  of  the  lump  burn.  So  for  this  experi- 
ment take  a  lump  of  quick  lime  out  of  a  recently  opened 
container.  Cover  it  with  water  in  a  porcelain  dish  and  then 
pour  off  the  excess  of  water  that  did  not  soak  into  the  porous 
lump.  Note  that  the  lump  soon  grows  very  hot  and  gives 
off  clouds  of  steam,  and  swells  up  and  tumbles  apart  to  form 
a  fluffy  white  powder.  Stir  some  of  this  powder  with  water. 
It  makes  a  milky  looking  suspension  which  will  turn  litmus 
blue.  If  some  of  this  suspension  is  placed  in  a  tall  bottle  the 
white  powder  settles  and  a  clear  liquid  remains  above.  This 
liquid  colors  litmus  blue.  Taste  it  by  taking  a  teaspoonful 


OXIDES  AND  WATER  63 

of  it  in  the  mouth;  note  the  taste  and  remember  that  it  is 
described  as  an  alkaline  taste.  The  liquid  is  called  lime 
water. 

It  is  obvious  from  the  large  amount  of  heat  developed  that 
the  residual  affinity  between  calcium  oxide  and  hydrogen  oxide 
is  very  great. 

CaO  +  H2O  ->  Ca(OH)2. 

The  same  substance  is  formed  here,  calcium  hydroxide,  as  is  ob- 
tained by  the  action  of  calcium  metal  on  water,  only  in  this  case 
no  hydrogen  is  displaced.  The  primary  valence  of  the  calcium 
is  already  satisfied  in  calcium  oxide  and  it  is  not  necessary  to 
dispossess  hydrogen  in  order  to  get  the  oxygen  or  hydroxide  radical 
to  satisfy  that  valence.  Calcium  hydroxide  is  not  very  soluble 
in  water  and  mainly  on  that  account  it  is  not  as  strong  a  base  as 
sodium  hydroxide. 

13.  Magnesium  Oxide  and  Water.    Burn  a  piece  of  mag- 
nesium ribbon  held  in  pincers  so  that  the  ash  falls  into  a 
clean  dish.     Stir  half  of  the  ash  into  a  beaker  full  of  water 
and  test  the  solution  with  litmus.     Wet  the  other  half  of  the 
ash  with  a  single  drop  of  water  and  place  the  moistened  mass 
on  one  side  of  strip  of  red  litmus  paper.     Look  on  the  other 
side  of  the  paper  and  note  that  in  a  little  while  the  center  of 
the  wet  spot  turns  blue. 

Magnesium  oxide  does  not  combine  as  energetically  with  water 
as  calcium  oxide  and  the  magnesium  hydroxide  formed  is  very- 
much  more  insoluble  than  calcium  hydroxide.  Thus  the  satu- 
rated solution  of  the  hydroxide  is  barely  alkaline  enough  to  color 
litmus  blue. 

The  hydroxides  of  aluminum  and  of  the  heavy  metals  are  much 
less  soluble  even  than  magnesium  hydroxide  and  for  the  most  part 
their  suspensions  do  not  affect  litmus.  The  hydroxides  of  the 
metals  are  however  considered  basic  although  very  weakly  so. 

14.  Non-metal  Oxides  and  Water.    Burn  small  amounts 
of  (a)  phosphorus   (use  red  phosphorus),   (6)   sulphur,  and 
(c)  carbon  (charcoal)  successively  in  large,  clean  bottles  of 
air.     The  phosphorus  and  sulphur  may  be  introduced  in  a 
deflagrating  spoon  made  by  winding  a  piece  of  chalk  with  a 


64  WATER  AND  SOLUTION 

piece  of  wire,  and  scooping  out  one  end,  but  the  charcoal, 
of  which  a  much  larger  piece  should  be  taken,  should  be 
attached  to  a  wire  and  held  in  the  flame  until  it  glows  brightly 
before  it  is  put  in  the  jar.  Quickly  add  3  to  5  cc.  of  water 
to  each  bottle,  close  it  with  a  glass  plate  (or  the  palm  of  the 
hand),  and  shake  vigorously  a  few  moments.  Test  the 
solutions  with  litmus.  Instead  of  preparing  the  carbon 
dioxide  solution  in  the  above  way  a  siphon  of  "  soda  water  " 
may  be  bought  at  a  drug  store.  In  this  bottle  water  has  been 
saturated  with  carbon  dioxide  under  a  high  pressure,  and 
consequently  there  is  a  good  deal  more  of  the  substance 
dissolved  and  its  properties  are  more  easily  detected.  Take 
a  little  of  the  soda  water  in  the  mouth  and  compare  its  taste 
with  that  of  lime  water.  This  has  an  acid  rather  than  an 
alkaline  taste. 

This  experiment  shows  that  oxides  of  non-metals  produce 
acids  and  also  that  the  acids  vary  a  good  deal  in  their  strength. 

P2O5  +  3  H2O  ->  2  H3PO4 
SO2  +  H2O  -> H2S03 
CO2  +  H2O  -*  H2C03 

The  oxides  of  phosphorus  and  sulphur  are  quite  readily  soluble 
but  the  carbon  dioxide  is  very  much  less  soluble. 

WATER  CONTAINS  Two  SEPARATELY  REPLACEABLE  PORTIONS 
OF  HYDROGEN 

In  the  formation  of  a  base  by  the  direct  action  of  a  metal  on 
water  hydrogen  was  displaced,  and  the  substance  sodium  hydrox- 
ide or  calcium  hydroxide  was  left.  The  formula  NaOH  or  Ca(OH)2 
shows  our  knowledge  that  the  compound  still  contains  one  half 
of  the  hydrogen  of  the  original  water,  but  let  us  think  ourselves 
for  a  moment  in  the  place  of  the  early  chemists  who  were  finding 
out  things  for  the  first  time.  They  carefully  dried  these  hydrox- 
ides and  then  tried  experiments  to  see  if  any  more  hydrogen  was 
obtained  from  the  hydroxides.  When  they  found  that  an  amount 
was  displaced  from  the  hydroxide  just  equal  to  that  displaced  by 
the  metal  from  the  water  when  the  hydroxide  was  prepared  they 
drew  the  conclusion  that  water  contained  two  parts  of  hydrogen 
in  combination  with  one  part  of  oxygen,  and  this  experiment  was 


CONCENTRATION  OF  SOLUTIONS  65 

one  of  the  strongest  arguments  which  led  to  the  adoption  of  the 
formula  H2O  for  water  rather  than  HO. 

15.  Mix  2  grams  of  dry  powdered  sodium  hydroxide  and 
3  grams  of  zinc  dust  in  a  mortar,  and  place  the  mixture  in 
a  hard  glass  test  tube  fitted  with  a  delivery  tube.  Heat  the 
mixture  in  the  tube  and  collect  a  little  of  the  evolved  gas 
and  show  by  a  test  that  it  is  hydrogen. 

Water  as  a  Solvent:   Concentration  of  Solutions 

The  importance  of  water  depends  in  very  large  measure  on  its 
ability  to  dissolve  other  substances.  Oftentimes  substances 
which  in  the  dry  state  will  not  react,  do  react  when  they  are 
dissolved  and  their  solutions  are  mixed.  If  a  reaction  is  to  be 
brought  about  between  two  substances  in  solution  it  is  a  matter 
of  importance  to  know  how  much  of  each  solution  to  take,  and 
to  do  this  it  is  necessary  to  know  the  concentration  of  the  solutions. 
The  following  definitions  make  Clear  the  methods  of  stating 
concentrations  and  the  problems  illustrate  the  advantages  of  the 
method. 

DEFINITIONS 

Mole.  A  mole  is  one  gram  molecular  weight  of  a  substance. 
e.g.  36.5  grams  of  HC1. 

Molal  Solution.  A  molal  solution  contains  one  mole  of  the 
dissolved  substance  in  one  liter  of  the  solution.  Note.  1  liter 
of  the  solution  contains  less  than  1000  grams  of  water,  but  the 
weight  of  the  whole  solution  is  usually  more  than  1000  grams. 

Equivalent.  An  equivalent  weight  of  a  substance  is  identical 
with  the  mole  when  the  valence  of  both  radicals  is  one.  e.g.  HC1, 
HNO3,  NaCl.  When  the  radicals  have  higher  valences,  the  equi- 
valent weight  contains  that  weight  of  each  radical  which  could 
either  combine  with  or  replace  1  gram  atomic  weight  of  hydrogen. 
Thus  the  moles  of  CuSO4,  H2SO4,  CuCl2  each  contain  2  equivalents. 

Normal  Solution.  A  normal  solution  contains  one  equivalent 
weight  of  the  dissolved  substance  in  1  liter  of  solution. 

Formula  Weight.  A  formula  weight  of  a  substance  means 
exactly  the  same  as  a  mole  if  the  formula  is  the  molecular  for- 
mula. Sometimes  however  the  molecular  weight  of  a  substance 
is  not  known  although  its  composition  is  known  and  an  empirical 


66  WATER  AND  SOLUTION 

formula  is  given.  In  such  a  case  the  term  mole  has  no  certain 
meaning,  but  the  meaning  of  formula  weight  is  perfectly  definite, 
it  is  the  number  of  grams  obtained  by  adding  up  the  total  of  the 
atomic  weights  in  the  formula  as  it  is  written. 

Formal  Solution..  A  formal  solution  contains  one  formula 
weight  of  the  dissolved  substance  in  1  liter  of  solution. 

Concentration.  The  concentration  of  a  solution  may  be  ex- 
pressed in  terms  of  the  molal,  normal,  or  formal  amounts  of  dis- 
solved substance.  Thus  10  liters  of  0.1  normal  HC1  contain  the 
same  amount  of  acid  as  1  liter  of  normal  or  as  100  cc.  of  10  normal 
HC1. 

PKOBLEMS 

1.  How  many  times  normal  is  each  of  the  following  solu- 
tions? 

(a)  molal  H2SO4. 
(6)  formal  A12(SO4)3. 

(c)  molal  A1C13. 

(d)  0.1  molal  Na3P04. 

(e)  3  molal  H2SO4. 
(/)  6molalHN03. 

(g)  2.5  formal  MgSO4. 

2.  How  many  tunes  formal  are  the  following  solutions? 

(a)  6  normal  HC1. 
(6)  Normal  Na3PO4. 

(c)  0.5  normal  K2SO4. 

(d)  0.1  normal  K4(FeC6N6). 

(e)  0.4  normal  Ba(OH)2. 
(/)  normal  CuSO4. 

3.  (a)  What  is  the  normality  of  a  solution  of  HC1  con- 
taining 39  percent  by  weight  of  HC1  and  of  spec.  grav.  1.19? 

(b)  of  a  solution  of  HC1  containing  20  percent  by  weight 
and  of  spec.  grav.  1.12? 

(c)  of  a   solution  of  HN03  containing   68.6   percent  by 
weight  and  of  spec.  grav.  1.41? 

(d)  of  sulphuric  acid  containing  96  percent  H2S04  and  of 
spec.  grav.  1.84? 

4.  What  is  the  percent  by  weight  of  the  following  solutions? 
(a)  6-normal  HN03,  spec.  grav.  1.195. 

(6)  6-normal  HC1  spec.  grav.  1.100. 


GENERAL  QUESTIONS  67 

(c)  6  normal  H2S04  spec.  grav.  1.181. 

(d)  0.5  normal  H(C2H3O2)  spec.  grav.  1.00. 

5.  How  many  cc.  of  the  concentrated  sulphuric  acid  of  3 
(d)  should  be  taken  to  make  one  liter  of  normal  solution? 

6.  How  many  liters  of  HC1  (gas,  figured  at  standard  condi- 
tions) are  required  to  make  1  liter  of  12  normal  acid? 

7.  (a)  How  many  cc.  of  6  normal  HC1  are  needed  to  dis- 
solve 1  gram  of  zinc.     Zn  +  2  HC1  -»  ZnCl2  +  H2? 

(6)  How  many  cc.  of  hydrogen  are  evolved? 

8.  (a)  What  weight  of  calcium  carbonate  will  react  with  1 
liter  of  6  normal  HC1?     CaC03  +  2  HC1  ->  CaCl2  +  H2O  + 
C02. 

(6)  What  volume  of  carbon  dioxide  will  be  evolved? 

9.  It  is  desired  to  find  the  concentration  of  a  solution  of 
calcium  hydroxide.     500  cc.  of  this  solution  are  carefully  mea- 
sured into  a  beaker,  litmus  is  added  and  a  0.5  normal  solution 
of  HC1  is  run  in  until  the  color  just  changes  from  blue  to  red. 
The  volume  of  the  HC1  solution  thus  used  is  40  cc. 

(a)  What  is  the  normal  concentration  of  the  calcium  hy- 
droxide solution? 

(6)  the  molal  concentration? 

(c)  the  percent  by  weight  (spec.  grav.  =  1.00)? 

GENERAL  QUESTIONS  II. 

1.  Define  Solution. 

2.  Water  is  essential  to  the  maintenance  of  plant  and 
animal  life.     Discuss  the  properties  of  water  that  make  it  so. 

3.  Discuss  the  nature  of  the  compounds  of  salts  and  water. 

4.  Describe  two   different  classes  of  compounds  formed 
from  oxides  and  water. 

5.  What  is  the  meaning  of  the  term  hydrate?     Why  is 
the  compound  of  an  oxide  and  water  not  regarded  as  a  hy- 
drate? 

6.  Explain   the   use   of   the   term    hydroxide   in   naming 
the  compound  of  a  metal  oxide  and  water. 

7.  Contrast  the  action  of  a  metal  with  that  of  a  metal 
oxide  on  water  in  the  formation  of  a  hydroxide. 

8.  Contrast  the  formation  of  chlorine  hydrate  with  the 
action  of  chlorine  and  water  in  sunlight. 


CHAPTER   III 
THE  THEORY   OF  IONIZATION 

When  a  substance  dissolves  in  a  liquid  it  becomes  dispersed 
just  as  a  gaseous  substance  is  dispersed.  It  behaves  in  every  way 
similarly  to  a  gas  except  that  the  space  between  its  molecules  is 
occupied  by  the  liquid  instead  of  being  vacant.  Indeed  the  dis- 
solved substance,  the  solute,  obeys  almost  quantitatively  the  gas 
laws  (see  Chapter  I). 

But  abnormality  is  observed  with  a  certain  class  of  dissolved 
substances  which  are  called  electrolytes.  Solutions  of  electro- 
lytes conduct  electricity  very  well,  although  the  pure  substance, 
and  the  pure  solvent  by  themselves  do  not  conduct.  Methods  of 
determining  the  number  of  molecules  present  in  solutions  of  electro- 
lytes —  freezing  point  method,  osmotic  pressure  method,  etc.  — 
show  more  molecules  to  be  present  than  could  be  accounted  for 
if  the  substance  existed  as  an  ordinary  gas.  These  facts  are  ac- 
counted for  by  the  Theory  of  lonization  according  to  which  elec- 
trolytes in  solution  are  dissociated  into  their  radicals,  and  further- 
more these  radicals  are  electrically  charged.  These  charged 
independent  radicals  are  free  to  move  in  opposite  directions  under 
electrical  attractions  and  thus  the  solutions  conduct  electricity. 
Since  by  the  act  of  ionization  an  ordinary  molecule  breaks 
up  into  two  (Na+Cl-)  three  (Ca++Cl-Cl-)  (K+K+SO4~) 
four  (Fe+++Cl-Cl-Cl-)  (Na+Na+Na+PO4— ~)  or  even  five 
(K+K+K+K+Fe(CN)6—  — )  charged  radicals,  or  ions,  each  of 
which  behaves  in  most  respects  like  an  independent  molecule,  it 
is  clear  that  the  theory  of  ionization  accounts  for  the  abnormally 
large  number  of  molecules  as  well  as  for  the  electrical  conductivity. 

A  careful  study  of  the  laboratory  experiments  and  a  working 
out  of  the  problems  in  the  following  sections  should  give  one  an 
understanding  of  the  theory  of  ionization,  a  theory  which  is  of 
the  greatest  service  in  interpreting  the  chemistry  of  solutions. 
Although  the  experiments  are  printed  consecutively  in  one  sec- 
tion and  the  problems  in  the  next,  the  two  sections  should  be 
studied  together. 

68 


ELECTRICAL  CONDUCTIVITY  69 

Experiments 

1.  Osmotic  Pressure.  The  formation  of  osmotic  membranes, 
as  well  as  the  existence  of  osmotic  pressure,  may  be  qualita- 
tively shown  by  what  may  be  called  the  mineral  garden,  pre- 
pared as  follows:  small  lumps  or  crystals  of  certain  very  solu- 
ble salts,  such  as  ferric  chloride,  copper  chloride,  nickel  nitrate, 
cobalt  chloride,  and  manganese  sulphate,  are  dropped  into  50  cc. 
of  a  solution  of  sodium  silicate,  or  water  glass  (sp.  gr.  1.1),  in  a 
small  beaker.  Success  of  the  experiment  depends  on  using  small 
fragments  of  the  salts.  Their  behavior  resembles  that  of  grow- 
ing seeds,  as  they  appear  to  sprout  immediately  and  to  send  up 
shoots  toward  the  surface  of  the  liquid,  which  grow  with  a  visible 
rapidity. 

In  fact,  the  salts  at  once  commence  to  dissolve,  forming  thin 
layers  of  very  concentrated  solution  about  each  lump.  At  the 
surface  separating  each  of  these  layers  of  solution  from  the  water 
glass,  there  forms  a  coherent  film  of  the  insoluble  silicate  of  the 
metal.  This  film  is  an  osmotic  membrane  which  allows  water  to 
pass  either  in  or  out;  but  the  molecules  of  salt,  not  being  able 
to  pass  through,  exert  against  it  their  osmotic  pressure,  and  cause 
it  to  stretch  or  even  break.  If  it  breaks,  a  new  surface  of  the 
salt  solution  is  exposed  to  the  sodium  silicate,  and  a  new  film 
forms.  Since  the  film  is  thinnest  at  the  top  the  growth  is  mainly 
at  the  top  and  thus  the  little  tubes  of  silicate  of  the  metal 
shoot  up  through  the  solution.  Clusters  of  these  tubes  of  various 
colors  give  an  appearance  of  plant  growth  within  the  liquid. 

Record  this  experiment  in  the  note  book,  giving  a  description 
and  an  explanation  in  your  own  words.  Wash  out  beaker  im- 
mediately after  the  experiment,  as  the  sodium  silicate  solution 
will  etch  the  glass. 

ELECTRICAL   CONDUCTIVITY  OF  SUBSTANCES  IN  SOLUTION: 

Use  the  conductivity  apparatus  found  on  ends  of  desks,  a 
diagram  of  which  is  given  in  Figure  14. 

Electrodes  A  consist  of  stout  copper  wires,  and  are  to  be  used 
in  testing  the  conductivity  of  solid  substances,  lumps  of  which  are 
held  with  the  fingers  in  such  a  way  as  to  come  into  contact  with 
each  electrode.  A  16-candle  power  lamp  may  be  used  in  the 
socket  above. 


70  THEORY  OF   IONIZATION 

Electrodes  B  consist  of  similar  copper  wires,  but  bent  far  enough 
apart  so  that  they  will  pass  into  the  two  arms  of  a  U-tube  8  mm. 
internal  diameter  and  7  cm.  high,  when  the  latter  is  raised  from 
underneath.  A  50-candle  power  lamp  (or,  somewhat  less  satis- 
factorily, a  32-candle  poweMamp)  should  be  used  with  these 
eleclrodes,  which  are  then  designed  to 
show  differences  in  conductivity  among 
good  conductors. 

Electrodes  C  consist  of  fine  platinum 
wires  supported  upon  glass  rods,  and  are 
to  be  used  with  a  16-candle  power  lamp. 
They  are  to  be  used  in  testing  the  con- 
ductivity of  liquids,  the  latter  to  be 
placed  in  a  3-inch  vial.  This  vial  may 
be  raised  until  the  electrodes  are  im- 
mersed in  the  liquid.  Before  testing  the 
conductivity  of  any  given  solution  rinse 
the  platinum  electrodes  with  pure  water  until  when  they  are 
held  for  10  seconds  or  more  immersed  in  distilled  water  no  sign 
of  a  glow  is  seen  in  the  lamp  filament.  To  illustrate  the  neces- 
sity for  this  precaution,  immerse  the  electrodes,  first  in  hydro- 
chloric acid  solution,  then,  without  rinsing,  immerse  them  in  pure 
water. 

2.  Electrical  Conductivity,  (a)  Pure  substances.  Note  that 
the  lamp  does  not  glow  when  air  fills  the  space  between  the  elec- 
trodes. Then  raise  successively  between  electrodes  C  distilled 
water,  alcohol,  pure  acetic  acid  (labeled  Glacial  Acetic  Acid) 
and  place  in  contact  with  electrodes  A  lumps  of  any  two  dry 
salts  found  in  the  laboratory,  for  example  common  salt,  NaCl, 
and  blue  vitriol,  CuS04.5H2O. 

What  general  statement  can  be  made  about  the  conductivity 
of  pure  non-metallic  substances,  gaseous,  liquid,  or  solid? 

(6)  Solutions  of  Electrolytes.  Test  for  conductivity  each  of 
the  following  solutions :  Crush  about  J  gram  of  each  of  the  solid 
salts  tested  in  a  to  a  powder  and  dissolve  each  in  10  cc.  of  water. 
Add  10  drops  of  glacial  acetic  acid  to  10  cc.  of  water.  Dilute  2 
cc.  of  each  of  the  laboratory  acids  (which  are  already  in  solution 
in  water)  with  10  cc.  of  water.  Dissolve  about  J  gram  each  of 
sodium  hydroxide  and  potassium  hydroxide  in  10  cc.  of  water. 
Dilute  2  cc.  of  ammonium  hydroxide  solution  with  10  cc.  of  water. 


EXPERIMENTS  71 

List  the  above  solutions  in  the  order  of  their  conductivity. 
What  classes  of  substances  conduct  when  in  solution?  What 
explanation  can  you  give  of  electrolytic  conductance  and  how 
can  you  account  for  the  fact  that  some  solutions  conduct  much 
better  than  others? 

(c)  Solutions  of  non  Electrolytes.  Dissolve  J  gram  sugar,  J 
gram  urea,  J  cc.  alcohol,  \  cc.  glycerine  each  in  10  cc.  of  water, 
and  test  the  conductivity  of  the  solutions. 

3.  Acids.     Test  the  conductivity  of  pure  tartaric  acid  (a  solid) 
and  of  pure  acetic  acid  (glacial  acetic  acid) . 

Then  test  the  conductivity  of  these  same  acids  diluted  with 
10  to  20  parts  of  water,  and  also  of  other  common  laboratory  acids 
diluted  with  water. 

Look  up  the  percent  of  ionization  in  0.1  normal  solution  of 
each  of  the  acids.  What  component  is  common  to  all  acids  and 
is  responsible  for  the  characteristic  properties  of  acids?  Name 
three  other  components  (i.  e.  species  of  ions  or  molecules)  in  any 
acid  solution. 

4.  Strong  and  Weak  Acids.     In  order  to  compare  the  strength 
of  acids  it  is  necessary  to  have  solutions  of  the  same  concentration. 
Prepare  50  cc.  each  of  0.1  normal  hydrochloric  and  acetic  acids 
by  diluting  5  cc.  of  acid  from  bottle  labeled   1-normal  with  45 
cc.  water.     Carefully  compare  the  conductivity  of  these  0.1  nor- 
mal acids  using  Electrodes  C.     Carefully  compare  the  intensity 
of  the  sour  or  acid  taste.     Quite  a  bit  of  experimenting  is  neces- 
sary to  find  just  the  amount  of  acid  to  take  and  how  long  to  hold 
it  on  the  tongue  in  order  to  get  a  fair  estimate  of  the  comparative 
acidity.     It  must  also  be  borne  in  mind  that  the  strong  acid  may 
partly  paralyze  the  nerves  of  the  tongue  for  a  short  time;  there- 
fore after  tasting  one  acid,  rinse  out  the  mouth  and  wait  a  short 
time  before  tasting  another.     Compare  the  effect  of  the  two  0.1 
normal  acids  on  blue  litmus. 

The  conductivity  of  the  two  acids  ought  to  be  proportional  to 
the  degree  of  ionization;  likewise  the  sour  taste,  which  is  the 
property  of  the  hydrogen  ion,  ought  likewise  to  be  proportional  to 
the  degree  of  ionization;  even  the  weaker  acid  contains  enough 
hydrogen  ions  to  completely  turn  the  color  of  the  very  sensitive 
litmus,  so  no  difference  is  shown  by  this  indicator. 

Make  a  tabulation  for  each  acid  of  the  number  of  grams  of 
the  four  components  present  in  1  liter  of  0.1  normal  solution: 


72  THEORY  OF  IONIZATION 

(1)  water,  (2)  un-ionized  acid,  (3)  hydrogen  ion,  (4)  acid  radical 
ion.  Arrange  the  tabulation  for  each  acid  somewhat  on  tho 
following  plan. 

COMPONENTS   IN.  1   LITER,  OF  0.1   NORMAL   ACETIC   ACID 
total  acetic  acid  0 . 1  X  60  =      6.0  grams 

water  1000  -  6    =  994 

ionized  acetic  acid = " 

un-ionized  acid =" " 

H+  ion = " 

C2H3O2-  ion ,...  = " 

5.  Bases.     Test  the  electrical  conductivity  of  a  dry  lump  of 
sodium  or  potassium  hydroxide  (do  not  handle  it  with  the  fingers) 
before  its  surface  has  become  wet  by  taking  moisture  from  the 
atmosphere.     Then  test  the  conductivity  of  dilute  solutions  of 
sodium,  potassium,  and  ammonium  hydroxides.     Rub  a  single 
drop  of  6-normal  NaOH  lightly  between  the  thumb  and  forefinger 
and  note  the  slippery  feeling.     Immediately  rinse  the  alkali  from 
the  fingers  since  it  would  very  soon  take  off  'the  skin.     Repeat  the 
experiment  with  a  drop  of  6-normal  KOH  and  again  with  a  drop 
of  6-normal  NH4OH. 

Look  up  the  percent  of  ionization  in  0.1  normal  solution  of 
each  of  the  bases.  What  component  is  common  to  all  bases  and 
is  responsible  for  the  characteristic  properties  of  bases?  Name 
three  other  components  in  any  base  solution. 

6.  Strong  and  Weak  Bases.     Repeat  in  every  detail  Experiment 
4  using  sodium  hydroxide  and  ammonium  hydroxide  as  typical 
of  a  strong  and  a  weak  base.     Make  similar  tabulations  of  the 
weight  of  each  of  the  components,  (1)  water,  (2)  un-ionized  base, 
(3)  metal  radical  ion  and  (4)  hydroxyl  ion,  in  one  liter  of  0.1  normal 
solution. 

7.  Neutralization  of  a  Strong  Acid  and  a  Strong  Base.     Fill 
a  narrow  U-tube  with  molal  hydrochloric  acid  and  insert  electrodes 
B  until  the  lamp  glows  dimly  but  distinctly.     Note  carefully  the 
depth  to  which  the  electrodes  are  inserted.     Then  refill  the  tube 
with  molal  sodium  hydroxide,  again  insert  the  electrodes  to  the 
same  depth,  and  notice  how  strongly  the  lamp  glows. 

Run  10  cc.  of  molal  hydrochloric  acid  into  a  beaker,  add  1  drop 
of  litmus  solution,  and  then  add  molal  sodium  hydroxide  until 
the  color  changes  to  blue.  Add  a  drop  or  two  more  acid  until  the 


EXPERIMENTS  73 

color  again  changes,  and  finally  bring  the  solution  to  the  exact 
neutral  point  when  1  drop  of  acid  will  turn  the  litmus  red  and  a 
single  drop  of  base  will  bring  back  the  blue. 

Fill  the  same  U-tube  with  this  neutralized  solution,  insert  the 
electrodes  to  the  same  depth  as  before,  and  compare  the  conduc- 
tivity of  the  neutral  solution  with  that  of  the  acid  and  the  base. 

Explain  the  process  of  neutralization  according  to  the  ionic 
theory  and  account  for  any  differences  noted  in  the  conductivity. 
Also  write  the  ionic  equation. 

Write  ionic  equations  from  now  on  whenever  it  is  possible.  Use 
the  intersecting  method  as  described  in  the  notes  in  the  second 
part  of  this  chapter. 

8.  Neutralization  of  a  Weak  Acid  and  a  Weak  Base.    Test 
separately  the  conductivity  of  molal  acetic  acid  and  of  molal 
ammonium  hydroxide,  this  time  using  electrodes  C.     Neutralize 
10  cc.  of  the  molal  acid  by  adding  the  molal  base  in  the  same 
manner  as  in  No.  7;  and  compare  the  conductivity  of  the  neutral- 
ized solution  with  that  of  the  acid  and  base  separately. 

Explain  the  neutralization  of  a  weak  acid  and  a  weak  base 
according  to  the  ionic  theory  and  account  for  the  change  observed 
in  the  conductivity.  Explain  why  the  neutral  solution  contains 
no  undissociated  molecules  of  the  acid  and  base. 

Remember  the  injunction  at  the  end  of  the  preceding  experiment 
to  write  ionic  equations  as  a  part  of  the  explanation  of  every  re- 
action. Do  not  however  neglect  to  give  the  explanation  in  words 
as  well  as  by  means  of  ionic  equations. 

9.  High  lonization  of  all  Salt  Solutions.     The  object  of  Ex- 
periments 7  and  8  was  to  show  the  relative  number  of  ions  in 
equivalent  acid  and  base  solutions  and  in  the  neutral  solution 
resulting  from  adding  the  two  together.     In  Experiment  7  the 
same  electrodes  held  the  same  distance  apart  with  the  same  vol- 
ume of  liquid  between  them  were  used  on  all  three  solutions  tested. 
Hence  the  brightness  of  the  lamp  was  proportional  to  the  number 
of  ions  in  the  solution.     But  in  Experiment  8  quite  a  different 
kind  of  electrodes  was  used,  and,  although  a  valid  comparison  of 
the  concentration  of  ions  in  the  three  solutions  of  that  experiment 
was  obtained,  the  brightness  of  the  lamp  shown  for  the  neutral 
solutions  of  the  two  experiments  gives  no  comparison  of  the  num- 
ber  of  ions  contained.     Remember  that  the  neutral  solution  in 
each  experiment  is  one  half  normal  with  respect  to  the  salt. 


74  THEORY  OF  IONIZATION 

Prepare  again  a  neutral  solution  of  sodium  chloride  by  neutral- 
izing normal  sodium  hydroxide  with  normal  hydrochloric  acid,  and 
a  neutral  solution  of  ammonium  acetate  by  neutralizing  normal 
ammonium  hydroxide  with  normal  acetic  acid.  Test  the  con- 
ductivity of  both  -solutions  .^with  electrodes  A  and  again  test 
both  with  electrodes  C. 

What  general  statement  can  be  made  about  the  ionization  of 
salts.  Explain  again  (although  the  write-up  of  Experiment  8 
should  already  contain  the  explanation)  why  the  salt  of  a  weak 
acid  and  a  weak  base  can  be  as  highly  ionized  as  the  salt  of  a 
strong  acid  and  a  strong  base. 

10.  Displacement  of  a  Weak  Acid  from  its  Neutral  Salt  by 
Means  of  a  Stronger  Acid,  (a)  Heat  2  cc.  of  a  solution  of  sodium 
acetate  to  boiling  in  a  test  tube  and  observe  if  there  is  any  odor. 
Add  a  few  drops  of  a  strong  acid  and  observe  if  the  odor  of  acetic 
acid  (vinegar  odor)  can  be  detected. 

How  does  this  experiment  illustrate  the  displacement  of  a  weak 
acid  from  its  salt?  Suppose  that  exactly  equal  volumes  of  0.2 
normal  NaC2H3O2  and  0.2  normal  HC1  had  been  used,  what  per- 
cent of  the  way  to  completion  would  the  reaction  have  taken 
place?  Compare  the  completeness  of  this  reaction  with  that  of 
neutralization. 

(6)  To  2  cc.  of  a  solution  of  sodium  sulphite  add  hydrochloric 
acid  and  observe  the  odor.  Compare  this  odor  with  that  of  the 
bottle  of  sulphurous  acid.  Is  this  reaction  more  or  less  complete 
than  the  one  in  a?  Sulphurous  acid  dissociates  non-electrolytic- 
ally 

H2S03  ?±  H20  +  S02 

and  sulphur  dioxide  is  expelled  from  a  solution  by  boiling.  What 
effect  would  boiling  the  solution  have  on  the  degree  of  displace- 
ment of  the  weak  acid? 

(c)  To  5  cc.  of  sodium  carbonate  solution  add  acetic  acid,  a 
few  drops  at  a  time,  till  action  ceases.     What  is  the  gas  formed? 
What  acid  is  displaced  in  this  case?     How  is  the  completeness  of 
its  displacement  affected  by  the  escape  of  the  gas?     If  the  gas 
had  been  prevented  from  escaping  by  holding  the  solution  under 
high  pressure,  state  what  components  the  solution  would  hold, 
and  roughly  the  relative  amounts  of  each. 

(d)  Treat  a  small  quantity  of  marble  dust  (calcium  carbonate) 


EXPERIMENTS  75 

with  an  excess  of  dilute  hydrochloric  acid,  and  see  if  it  will  all 
dissolve.  Calcium  carbonate  usually  passes  as  an  insoluble  sub- 
stance, nevertheless  it  is  slightly  soluble  in  pure  water.  Look 
up  its  exact  solubility.  How  would  the  acid  react  with  the  small 
amount  in  solution?  How  would  the  equilibrium  between  the 
solid  calcium  carbonate  and  its  saturated  solution  be  affected 
by  this  reaction?  To  sum  up,  explain  how  the  calcium  carbonate 
dissolves  in  acids. 

11.  The  Displacement  of  a  Weak  Base  from  its  Neutral  Salt 
by  Means  of  a  Stronger  Base.     Warm  2  cc.  of  ammonium  chlo- 
ride solution,  and  observe  if  there  is  an  odor.     Add  2  cc.  of  so- 
dium hydroxide  solution,  and  again  observe  if  there  is  an  odor. 

The  odor  is  that  of  the  gas  NH3  and  it  indicates  the  presence  of 
ammonium  hydroxide  in  the  solution.  If  equal  volumes  of  cold 
0.2  normal  solutions  of  NH4C1  and  NaOH  are  mixed  to  what 
degree  of  completion  does  the  reaction  proceed?  The  amount  of 
NH3  gas  escaping  from  such  a  cold  solution  is  negligible.  If  the 
solution  is  boiled,  facilitating  the  escape  of  NH3  what  will  be  the 
final  result? 

12.  The  Displacement  of  an  Insoluble  Base  from  its  Neutral 
Salt  by  Means  of  a  Strong  Base.     To  2  cc.  of  magnesium  sulphate 
solution  add  a  little  sodium  hydroxide  solution.     Look  up  the 
solubility  and  the  degree  of  ionization  of  magnesium  hydroxide. 

CHARACTERISTIC  REACTIONS  OF  CERTAIN  IONS 

If  two  ionized  substances  are  brought  together  in  a  solution, 
and  one  of  the  possible  new  combinations  of  a  positive  and  a 
negative  ion  is  an  insoluble  solid  substance,  that  substance  will 
form  instantly  as  a  precipitate.  Characteristic  precipitates  serve 
as  a  means  of  identifying  specific  ions. 

Some  ions  possess  characteristic  colors  which  they  impart  to  a 
clear  solution.  Thus  the  cupric  ion,  Cu++,  is  blue  and  all  solu- 
tions of  cupric  salts  are  clear  blue  unless  the  color  is  modified  by 
another  colored  component.  Thus  the  appearance  of  a  color  or 
a  change  of  color  in  a  clear  solution  when  a  reagent  is  added  helps 
to  identify  the  ions  present. 

In  the  following  experiments  an  inexperienced  student  always 
is  impelled  to  make  the  mistake  of  using  too  concentrated  solu- 
tions and  adding  too  much  reagent.  This  not  only  wastes  ex- 
pensive chemicals,  but  it  obscures  the  effects  to  be  observed, 


76  THEORY  OF  IONIZATION 

The  procedure  that  should  be  followed  is  to  take  about  1  cc.  of 
the  solution  to  be  tested,  dilute  this  with  5  cc.  of  water,  and  add 
the  reagent  a  single  drop  at  a  time,  shaking  thoroughly  after  each 
drop.  In  this  way  keep  adding  reagent  until  no  further  change 
is  brought  about.  It  often  happens  that  a  limited  amount  of 
reagent  will  produce  an  effect,  say  a  precipitate,  and  a  larger 
amount  will  produce  another  effect,  say  redissolve  the  precipitate. 
If  the  reagent  is  "  dumped  in  "  carelessly  these  effects  may  be 
missed. 

13.  Chloride  Ions.     To  solutions  of  various  chlorides,  say  hy- 
drochloric acid,  sodium  chloride,  and  calcium  chloride,  add  a  few 
drops  of  a  solution  of  a  silver  salt  —  silver  nitrate  or  silver  sul- 
phate. 

To  a  solution  of  potassium  chlorate  (be  sure  that  it  is  free  from 
chloride)  add  a  few  drops  of  silver  nitrate. 

How  may  the  presence  of  chloride  ions  be  recognized?  Why 
is  not  the  same  test  given  by  the  chlorine  in  potassium  chlorate? 

14.  Sulphate   Ions.    To   solutions   of   soluble   sulphates,   say 
sodium  sulphate,  copper  sulphate,  sulphuric  acid,  add  barium 
chloride  solution.     After  the  effect  of  this  reagent  is  noted  add 
hydrochloric  acid  to  see  if  the  precipitate  is  redissolved  by  acid. 

Repeat  the  experiment  using  lead  nitrate  instead  of  barium 
chloride  as  the  reagent. 

15.  Copper  Ions,     (a)  To  a  solution  of  cupric  salt,  say  CuSO4, 
add  NH4OH  in  small  amount  and  then  in  excess. 

In  writing  ionic  equations  consider  the  light  blue  precipitate 
as  Cu(OH)2,  for  the  sake  of  simplicity,  instead  of  the  rather  in- 
definite basic  salt.  To  explain  the  deep  blue  color  see  note  on 
Ammoniates  later  in  this  chapter. 

(6)  To  another  sample  of  cupric  salt  solution  add  ammonium 
sulphide.  Divide  the  black  suspension  in  two  parts,  and  add  HC1 
in  excess  to  one  part,  and  NH4OH  in  excess  to  the  other,  to  see  if 
the  precipitate  is  soluble  in  either  of  these  reagents. 

16.  Zinc  Ions.     Repeat  every  step  of  the  preceding  experiment 
using  a  zinc  salt  instead  of  a  copper  salt. 

Using  the  information  gained  in  these  two  experiments,  devise 
a  method  by  which  you  could  demonstrate  the  presence  of  a  small 
amount  of  zinc  salt  in  a  solution  containing  copper  salt. 

17.  Ferrous  Ions.     Repeat  a  and  b  of  Experiment  15  upon  a 
ferrous  salt  instead  of  a  cupric  salt. 


EXPERIMENTS  77 

(c)  to  a  solution  of  ferrous  sulphate  add  potassium  ferricyanide 
K3(Fe(CN)6). 

18.  Ferric  Ions,     (a)  To  a  solution  of  ferric  salt  say  FeClg  add 
NH4OH. 

(6)  To  other  samples  of  the  ferric  salt  solution  add  potassium 
ferrocyanide  K4(Fe(CN)6)  and  potassium  thiocyanate,  KCNS, 
respectively. 

19.  Silver    Ions,     (a)    Recalling   Experiment    13    state    how 
chloride  ions  may  be  used  as  a  reagent  for  silver  ions. 

(b)  To  2  cc.  of  ^-normal  silver  nitrate  solution  add  NH4OH 
very  carefully.     Even  a  few  drops  of  the  6-normal  reagent  will 
prove  an  excess.     So  dilute  1  cc.  of  the  reagent  with  10  cc.  of  water 
in  another  test  tube,  dip  a  stirring  rod  in  this  solution  and  carry 
it,  hanging  to  the  rod,  a  drop  at  a  time  to  the  AgNO3  solution. 

Find  in  a  reference  book  the  formula  of  the  ammonio-silver 
ion. 

20.  Electromotive  Series  of  the  Metals,     (a)     Place  a  few 
pieces  of  zinc  in  5  cc.  of  ^-normal  solution  of  copper  sulphate, 
shake  the  mixture  frequently,  and  after  about  15  minutes  with- 
draw 1  cc.  of  the  solution  and  test  it  for  copper  and  for  zinc 
ions,  applying  the  information  obtained  from  the  preceding  ex- 
periments.    If  any  copper  ions  are  still  present  let  the  remainder 
of  the  mixture  stand  for  15  minutes  more  with  frequent  shaking, 
and  repeat  the  test.     Continue  until  you  have  reached  a  conclu- 
sion as  to  whether  copper  ions  can  be  completely  displaced  from 
solution  by  zinc. 

(6)  Repeat  (a)  using  5  cc.  of  about  0.2  molal  CuSO4  solution 
and  some  clean  pieces  of  iron  wire. 

(c)  Again  repeat  (a)  using  5  cc.  of  y^-normal  AgNOs  solution 
and  some  clean  pieces  of  copper  wire. 

(d)  Recall  (or  find  out  by  experiment,  if  preferred)  the  be- 
havior of  silver,  copper,  iron  and  zinc  with  hydrochloric  acid  or 
dilute  sulphuric  acid.     Make  a  list  of  these  metals  including  hy- 
drogen in  the  order  of  their  ionizing  potential. 

21.  Hydrolysis.     Dissolve  about  half  a  gram  each  of  ferric 
chloride,  sodium  chloride,  and  sodium  carbonate  in  a  little  water, 
and  test  each  solution  with  red  and  blue  litmus. 

Explain  the  relation  of  hydrolysis  to  the  observed  results. 
Are   the   reactions  of  hydrolysis    complete;     if   not,   explain 
why. 


78  THEORY  OF  IONIZATION 

22.  Hydrolysis  a  Reversible  Reaction.     Place  |  gram  of  solid 
zinc  chloride  in  a  dry  test  tube  and  add  water  a  single  drop  at  a 
time  with  constant  shaking  until  the  solid  dissolves  giving  a  clear 
sirupy  solution.     Then  fill  the  tube  half  full  with  water  and  ob- 
serve the  flocculent  precipitate^     Now  add  hydrochloric  acid,  drop 
by  drop,  till  the  precipitate  disappears. 

Assuming  the  precipitate  to  have  been  basic  zinc   chloride, 

/OH 

Zn<;  explain   its  formation  from   the'-  components   of  the 

xci, 

solution,  and  explain  why  a  slight  excess  of  hydrochloric  acid 
causes  it  to  disappear. 

23.  Solubility  Product.     Recall  that  a  saturated  solution  can 
be  prepared  by  stirring  an  excess  of  the  powdered  substance  for 
a  very  long  time  with  water.     The  undissolved  material  can  be 
allowed  to  settle  and  the  clear  saturated  solution  may  be  de- 
canted off  or  poured  through  a  filter.     Note  however  that  the 
solubility  of  silver  acetate  and  potassium  chlorate,  the  salts  used 
in  this  experiment,  varies  very  greatly  with  the  temperature.     If 
a  saturated  solution  were  prepared  one  day  in  a  stock  bottle  and 
the  bottle  allowed  to  stand  on  the  reagent  shelf  over  night,  at 
the  lower  temperature  during  the  night  the  solution  would  be- 
come supersaturated  and  crystals  of  the  salt  would  separate  out. 
In  the  morning  when  it  warmed  up  again  the  solution  would 
be  very  much  undersaturated,  and  it  would  take  prolonged  stir- 
ring to  get  it  again  saturated.     Hence  it  is  useless  to  perform 
this  experiment  unless  one  makes  certain  that  he  really  has  a 
saturated  solution  to  start  with. 

(a)  Place  2  cc.  of  a  saturated  solution  of  silver  acetate  in  each 
of  three  test  tubes.  Add  to  one  tube  a  small  crystal  of  silver 
nitrate  (not  more  than  0.05  gram) .  Agitate  the  solution  until  the 
crystal  has  completely  dissolved.  If  no  change  is  noticed  at 
once,  set  the  tube  aside  for  several  minutes  and  observe  again. 

To  the  second  tube  add  a  small  crystal  of  sodium  acetate  in- 
stead of  silver  nitrate,  and  to  the  third  tube  add  a  crystal  of  am- 
monium nitrate  somewhat  larger  than  either  of  the  crystals  already 
used.  The  ammonium  nitrate  should  have  been  tested  to  see 
that  it  contains  no  chloride. 

Silver  acetate  is  only  slightly  soluble,  that  is,  about  10  grams 
per  liter  at  20°  C. 


NOTES  AND  PROBLEMS  79 

Put  all  the  waste  from  this  experiment  in  the  bottle  marked 
silver  residues. 

(6)  Repeat  the  above  experiment,  substituting  a  saturated 
solution  of  potassium  chlorate  and  crystals  of  potassium  chloride, 
sodium  nitrate,  and  sodium  chlorate. 

24.  Effect  of  Its  Neutral  Salt  on  Strength  of  a  Weak  Acid. 
(a)  Methyl  orange  is  an  indicator  which  is  used  much  in  the 
same  way  as  litmus  to  show  the  presence  of  hydrogen  or  hydroxyl 
ions.  In  the  presence  of  hydrogen  ions  it  is  pink,  and  of  hydroxyl 
ions  yellow;  with  varying  very  small  concentrations  of  hydrogen 
ions  it  shows  intermediate  colors. 

In  four  test  tubes  place  respectively  10  cc.  normal  HC1,  10  cc. 
normal  HAc,  10  cc.  normal  HAc,  10  cc.  normal  NaOH,  and  to 
each  add  two  drops  of  methyl  orange  solution.  The  first  tube 
shows  the  pink  color  imparted  by  strong  acids  and  the  fourth  the 
yellow  color  imparted  by  bases.  The  second  and  third  tubes 
should  of  course  show  exactly  the  same  color  tone.  Now  drop 
into  the  third  tube  about  a  gram  and  a  half  of  crystallized  sodium 
acetate,  NaC2H3O2.3H2O  and  dissolve  the  salt  by  shaking.  Now 
compare  the  colors  of  the  second  and  third  tubes,  still  keeping 
the  first  and  fourth  tubes  for  reference  colors. 

(b)  The  acid  strength  may  be  compared  by  the  rapidity  of  the 
reaction  with  calcium  carbonate.  In  each  of  two  small  beakers 
place  1  gram  of  powdered  calcium  carbonate  and  10  cc.  of  water 
and  stir  until  the  powder  is  entirely  wet.  Have  ready  two  solu- 
tions as  follows:  one  consists  of  20  cc.  of  normal  acetic  acid,  the 
other  of  20  cc.  of  normal  acetic  acid  in  which  an  equivalent  amount, 
about  3  grams  of  sodium  acetate  NaC2H3O2.3H20  has  been  dis- 
solved. At  the  same  instant  add  the  two  solutions  to  the 
respective  beakers  containing  calcium  carbonate.  The  rapidity  of 
effervescence  should  be  compared,  also  the  time  it  takes  for  the 
solid  to  entirely  dissolve. 

Notes  and  Problems 
MEASUREMENT  OF  IONIZATION 

Molal  Lowering  of  the  Freezing  Point.  Dissolved  substances 
have  the  effect  of  lowering  the  temperature  at  which  the  solvent 
freezes,  and  a  very  remarkable  law  is  found  to  hold  in  regard  to 
the  extent  of  this  effect,  namely: 


80  THEORY   OF  IONIZATION 

Equimolal  amounts  of  all  dissolved  substances  whose  solutions 
do  not  conduct  electricity  have  the  same  effect  upon  the  freezing 
point  of  the  solvent,  irrespective  of  the  weight  in  grams  or  the 
chemical  nature  of  the  substance. 

The  freezing  point  lowering,  is  proportional  to  the  number  of 
moles  of  solute  in  a  given  weight  of  solvent,  and  for  one  mole  of 
solute  in  1000  grams  of  water  the  lowering  is  1.86  degrees  Centi- 
grade. This  figure  is  known  as  the  molecular  lowering  of  the  freez- 
ing point  for  water. 

1.  At  what  temperature  will  a  solution  freeze  that  is  made 
by  dissolving  1  gram  of  sugar,  C^H^On,  in  10  grams  of  water? 

2.  One  mole  of  urea,  CO(NH2)2,  in  10  liters  of  water? 

3.  How  many  grams  of  methyl  alcohol,  CH3OH,  should  be 
added  to  1000  grams  of  water  to  give  a  solution  that  will 
freeze  at  -  10°? 

4.  What  is  the  molecular  weight  of  a  substance,  3  grams  of 
which  dissolved  in  50  grams  of  water  gives  a  solution  freezing 
at  -  0.93°? 

5.  Pure  benzene  freezes  at  5.48°.     What  is  the  molecular 
lowering  of  the  freezing  point  for  benzene  if  a  solution  of 
6.4  grams  of  naphthalene,  Ci0H8,  in  100  grams  of  benzene 
freezes  at  3.03°? 

6.  When  4.88  grams  of  a  certain  other  substance  are  dis- 
solved in  50  grams  of  benzene  the  solution  freezes  at  2.85°. 
What  is  the  molecular  weight  of  the  substance? 

Osmotic  Pressure.  When  a  substance  dissolves  in  a  solvent  its 
molecules  lose  their  cohesion  for  each  other  just  as  when  a  solid 
or  liquid  substance  is  changed  to  a  gas.  The  dissolved  state  is 
like  the  gaseous  state,  in  that  the  molecules  of  the  dissolved  sub- 
stance are  widely  separated;  it  differs,  in  that  the  space  between 
the  molecules  is  filled  with  the  solvent. 

A  dissolved  substance  will  diffuse  so  as  to  make  its  concen- 
tration uniform  throughout  the  solution,  just  as  a  gas  will  expand 
to  fill  uniformly  the  vessel  in  which  it  is  confined.  The  pressure 
with  which  dissolved  substances  tend  to  diffuse  from  a  part  of 
the  solution  at  any  given  concentration  into  an  adjacent  portion 
of  pure  solvent  is  called  osmotic  pressure,  and  this  pressure  is  not 
only  of  the  same  nature  as  gas  pressure,  but  it  also  obeys  the  same 
laws,  —  Boyle's,  Charles',  and  Avogadro's  laws.  One  mole  of  any 


NOTES  AND  PROBLEMS  81 

non-electrolyte  in  22.4  liters  of  water  at  0°  has  an  osmotic  pres- 
sure of  1  atmosphere  —  the  same  pressure  as  1  mole  of  a  gas 
exerts  at  the  same  temperature  and  volume. 

Of  course  the  osmotic  pressure  is  manifest  only  between  different 
points  within  the  body  of  liquid  in  question.  The  dissolved  sub- 
stance not  being  able  to  escape  beyond  the  surface  of  the  solution, 
because  of  the  cohesive  power  of  the  whole  solution,  osmotic 
pressure  will  obviously  not  be  exerted  against  the  glass  walls  of  a 
containing  beaker,  or  against  the  air  above  the  solution. 

7.  Find  the  osmotic  pressure  of  1  gram  of  sugar  dissolved 
in  10  grams  of  water  at  0°. 

8.  -—of  1  gram  of  urea,  CO(NH2)2,  in  10  grams  of  water 
atO°. 

9.  How  many  grams  of  sugar  should  be  dissolved  in  1000 
grams  of  water  to  give  an  osmotic  pressure  of  1  atmosphere 
atO°? 

10.  —  to  give  the  same  pressure  at  38°? 

11.  What  is  the  molecular  weight  of  a  substance,  5  grams 
of  which  at  0°  in  250  cc.  of  water  has  an  osmotic  pressure  of 
2.24  atmospheres? 

lonization.  The  ions  of  a  substance  in  solution  exert,  of  course, 
enormous  attraction  for  ions  of  opposite  charge,  but  since  there 
are  ions  of  opposite  charge  lying  in  every  direction  from  each 
individual  ion,  the  effect  of  this  electrostatic  attraction  is  nullified 
and  each  ion  exists  as  a  freely  moving  independent  unit  in  the 
solution.  As  such  it  should  exert  the  same  osmotic  pressure  and 
have  the  same  effect  upon  the  freezing  point  as  any  complete  un- 
ionized molecule. 

12.  Assuming  complete  ionization  of  the  electrolyte,  what 
would  be  the  freezing  point  of  10  grams  of  NaCl  in  500  grams 
of  water? 

13.  —  of  10  grams  of  CaC^  in  500  grams  of  water? 

14.  —  of  10  grams  of  Feds  in  500  grams  of  water? 

15.  —  of  10  grams  of  K4Fe(CN)G  in  500  grams  of  water? 

K4Fe(CN)6  ->  4K+  +  Fe(CN)6~ 

16-19.    What  is  the  osmotic  pressure  at  0°  of  the  salt  in 
Questions  12-15,  inclusive? 


82  THEORY  OF  IONIZATION 

20.  What  is  the  osmotic  pressure  of  0.1  mole  of  NaCl  in 
1000  grams  of  water  if  the  salt  is  86  percent  ionized? 

21.  —  of  0.1  mole  of  BaCl2  in  1000  grams  of  water  if  the 
salt  is  72  percent  ionized? 

The  degree  of  ionfzation  of  ,a  salt  may  be  calculated  from  the 
freezing  point  of  its  solution.  For  example,  if  0.1  mole  of  K2SO4 
(17.4  grams)  dissolved  in  1000  grams  of  water  freezes  at  —0.454°, 
we  know  that  0.1  mole  of  un-ionized  substance  will  lower  the 
freezing  point  0.1  X  1.86  =  0.186°,  and  that  0.1  mole  of  a  com- 
pletely ionized  substance  giving  3  ions  will  lower  the  freezing 
point  3  X  0.186  =  0.558°.  The  actual  freezing  point  lowering, 
0.454°,  lies  between  these  values  and  thus  indicates  incomplete 
ionization.  The  proportion  of  the  salt  ionized  is  found  by  di- 
viding the  difference  caused  by  the  actual  ionization,  0.454  —  0. 186 
=  0.268°,  by  that  which  would  be  caused  by  100  percent  ion- 
ization, 0.558  -  0.186  =  0.372°,  thus  giving  0.268/0.372  =  0.72. 
Thus  the  salt  is  72  percent  ionized. 

22.  If  9.45  grams  of  chloracetic  acid,  H(C2H2O2C1),  dis- 
solved in  1000  grams  of  water  show  an  osmotic  pressure  of 
2.51  atmospheres,  find  the  percent  ionization  of  the  acid, 
assuming  that  only  one  hydrogen  is  ionizable  and  that  the 
negative  ion  has  the  composition  of  the  radical  shown  in  the 
parenthesis. 

23.  A  solution  of  101  grams  of  potassium  nitrate  in  1000 
grams  of  water  freezes  at  —3.05.     Calculate  what  percent 
of  the  salt  is  ionized. 

EXTENT   OF   IONIZATION 

The  solvent  is  undeniably  the  agent  which  causes  ionization. 
Hence  the  more  solvent,  in  other  words,  the  more  dilute  the  solu- 
tions, the  greater  the  percent  of  the  molecules  that  will  be  ionized. 
In  very  dilute  solutions  (1  mole  of  solute  in  10,000  liters  of  solvent) 
strong  electrolytes  are  practically  completely  ionized.  In  con- 
centrated solutions  they  are  less  "ionized;  for  example,  hydrochloric 
acid  in  12-normal  solution  is  only  13  percent  ionized. 

In  the  following  table  are  given  the  degrees  to  which  various 
typical  electrolytes  are  ionized;  it  will  be  noticed  that  the  figures 
given  represent  the  percent  of  the  solute  that  is  ionized  when  the 
solute  is  in  0.1  equivalent  solution.  It  must  be  remembered  that 


EXTENT  OF  IONIZATION 


83 


in  more  concentrated  solutions  this  percent  is  always  smaller, 
and  in  more  dilute  solutions  it  becomes  larger  and  approaches 
100  percent  as  a  limit  at  great  dilution. 


SALTS 


Neutral  salts,  with  very  few  exceptions,  are  highly  ionized.  If 
they  are  classified,  as  in  the  table  below,  according  to  the  valence 
of  their  ions,  it  is  found  that  all  belonging  to  any  one  class  have  prac- 
tically the  same  degree  of  ionization.  The  six  classes  which  are 
indicated  may  be  typified  by  KNO3,  Ba(NO3)2,  FeCl3,  K,[Fe(CN)«], 
K2SO4,  and  ZnSO4,  respectively. 


Type  of  salt 


Percentage  ionization  in 
0.1  equivalent  solution 


R+R- 

R++R-R- 
R+R+R—  . . . . 
R+++R-R-R-. 

R+R+R+R 

R++R—  . , 


86 
72 
72 
65 
65 
45 


ACIDS 


Substance 

Percentage  ionization  in 
0.1  equivalent  solution 

HC1,  HBr,  HI,  HNO3    ) 

90 

HC1O3,  HC1O4,  HMnCM  

TT  O/-A     ^  TT+     1     TTQO  — 

90 

HSO4~  +±  H+  +  SO4  — 

30 

H2C2O4  +±  H+  +  HC2O4- 

50 

TT   Qf\      >   TT-i-      1—    TTQ/"V   — 

20 

TTOQ  —  ->  jj-j-TL  gn  —  — 

1 

H3PO4  +±  H+  +  H2PO4~ 

27. 

H2PO4~  *±  H+  +  HPO4  —   

0.2 

HPO4  —  +±  H+  -h  PO4  

0.0002 

H3AsO4  +±  H+  +  H2AsO4~         

20 

HF                                               

9 

HNO2                                              

7 

HC2H3O2  ?±  H+  +  C2H3O2~  

1.4 

H2CO3  ?±  H+  +  HCO3~        

0.12 

HCO3~  ?±  H+  +  CO3  —  

0.002 

0.05 

HS~  ^  H+  +  S  —  

0.0002 

0.01 

TT    T)f"V       v    TT  1          i        TT    T>/"V    — 

0.01 

84  THEORY  OF  IONIZATION 

BASES 


Percentage  lomzation  in 
Substance 

0.1  equivalent  solution 


KOH,  NaOH.. 

Ba(OH)2  <=*  Ba++  +  2  OH~ . 

NH4OH.. 


86 
75 


1.4 


Ca(OH)2,  Mg(OH)2  are  but  slightly  soluble,  but  so  far  as  they  do 
dissolve  they  are  dissociated  to  about  the  same  extent  as  Ba(OH)2 
in  a  solution  of  the  same  concentration. 

The  hydroxides  of  the  heavy  metals  are  very  insoluble  and,  as  a 
rule,  very  weakly  basic. 

AgOH  is  soluble  to  the  extent  of  1  part  in  15,000  of  water,  in  which 
solution  about  33  percent  of  its  molecules  are  ionized.  It  is  thus 
a  moderately  strong  base. 

Hydroxides  of  the  type  Zn(OH)2,  Fe(OH)2,  Mn(OH)2  are  much 
less  basic  than  AgOH,  while  hydroxides  of  the  type  Fe(OH)3,  Cr(OH)3, 
Al(OH)s  are  least  basic  of  all. 

Pure  water  contains  0.0000001  mole  of  H+  ions  and  0.0000001  mole 
of  OH~  ions  per  liter. 


IONIC  REACTIONS 

It  is  very  probably  true  that  all  chemical  reactions  are  due  to  a 
rearrangement  of  the  electrical  forces  residing  in  the  atoms  and 
molecules  concerned.  But  this  is  more  or  less  in  the  realm  of 
speculation.  Reactions  taking  place  in  water  solution  among 
substances  that  are  ionized  are  clearly  electrical  in  nature.  On 
the  other  hand,  a  very  large  number  of  organic  substances  (that 
is,  compounds  of  carbon)  are  not  perceptibly  ionized  even  in 
water  solution.  At  ordinary  temperature  such  substances  do  not 
enter  into  rapid  chemical  reaction  as  electrolytes  do. 

Non-Ionic  Reactions.  This  term  is  applied  to  all  slow  reactions 
in  which  none  of  the  substances  involved  show  any  measurable 
degree  of  ionization.  These  reactions  conform  to  the  law  of  defi- 
nite proportions  and  to  all  the  ordinary  principles  of  chemistry. 
According  to  the  present  state  of  our  knowledge  we  class  these 
reactions  as  non-ionic,  and  for  most  purposes  there  seems  to  be 
no  great  advantage  in  trying  to  deal  with  them  from  the  electrical 
viewpoint. 

Ionic  Reactions.  All  reactions  among  ionized  substances  clearly 
involve  electrical  forces,  and  it  is  the  purpose  of  this  outline  to 
deal  with  various  types  of  ionic  reactions. 


IONIC  REACTIONS  85 

lonization  as  a  Reversible  Reaction.  The  ionization  of  a  sub- 
stance is  itself  a  chemical  reaction,  since  if  a  pure  substance,  for 
example,  sodium  chloride,  which  is  of  course  un-ionized,  is  dis- 
solved in  water  it  becomes  largely  changed  into  what  we  may 
justly  regard  as  two  new  substances,  NaCl  — >  Na+  +  Cl~.  We 
know  (see  table  of  ionization  values)  that  if  0.1  mole  NaCl  is 
dissolved  in  1  liter  of  water,  86  percent  becomes  changed  to  ions. 
In  other  words,  the  reaction  just  written  proceeds  86  percent  of 
the  way  towards  completion.  We  have  also  learned  that  if 
0.1  mole  of  NaCl  is  dissolved  in  10,000  liters  of  water  the  ioniza- 
tion becomes  practically  complete.  In  other  words,  the  reaction 
has  proceeded  100  percent  of  the  way  towards  completion.  If 
now  this  very  dilute  solution  is  evaporated  until  but  1  liter  of 
water  is  left,  we  have  the  0.1  mole  of  NaCl  again  in  1  liter,  and 
at  this  concentration  we  know  it  is  but  86  percent  ionized.  In 
other  words,  the  reaction  of  association  (the  opposite  of  ioniza- 
tion) Na+  +  Cl~  — >  NaCl  must  have  progressed  14  percent 
of  the  way  towards  completion.  Obviously,  this  reaction  is 
reversible,  and  such  a  condition  is  indicated  m  an  equation  by 
employing  arrows  pointing  in  both  directions : 

NaCl  ±5  Na+  +  C1-. 

Equilibrium.  When  0.1  mole  of  NaCl  is  dissolved  in  1  liter  of 
water  and  the  two  opposing  reactions  have  adjusted  themselves 
so  that  86  percent  of  the  salt  is  in  the  ionized  condition  and  14 
percent  is  in  the  un-ionized  condition,  a  state  of  equilibrium  is 
said  to  exist.  Such  a  point  of  equilibrium  exists  for  every  ionic 
reaction  and,  as  in  the  case  just  cited,  it  is  independent  of  the 
direction  from  which  it  is  approached. 

lonization  is  a  very  rapid  reaction,  as  is,  indeed,  the  reverse 
reaction,  or  association;  and  the  point  of  equilibrium  is  reached 
with  great  rapidity.  In  fact,  in  the  time  taken  to  dissolve  a  salt 
in  water,  and  uniformly  mix  the  solution  by  stirring,  complete 
equilibrium  is  attained. 

For  the  sake  of  comparison  a  non-ionic  reaction  may  be  cited. 
If  two  volumes  of  hydrogen  (uncombined  hydrogen,  not  hydrogen 
ions)  are  mixed  with  one  volume  of  oxygen  in  a  glass  jar  at  ordi- 
nary temperature,  nothing  appears  to  happen.  Yet  we  know  this 
mixture  is  not  in  equilibrium.  By  the  careful  use  of  catalyzers,  the 
two  gases  will  combine  completely  to  form  water,  even  if  the  tem- 


86  THEORY  OF  IONIZATION 

perature  is  kept  from  rising.  As  we  well  know,  if  a  spark  is  applied 
to  the  mixture  a  violent  reaction  takes  place.  Equilibrium  for  this 
reaction  exists  only  when  it  has  gone  practically  to  completion  in 
the  direction  2H2  +  O2  — >  2H2O,  and  yet  at  ordinary  temperature 
and  without  catalyzers  the  reason  is  so  slow  that  it  will  not  have 
reached  an  equilibrium  condition  in  a  many  years. 

Equations  for  Ionic  Reactions.  Reactions  involving  ionized 
substances  cannot  be  adequately  represented. by  single  equations, 
because  such  equations  cannot  show  all  the  species  of  ions  and 
molecules  that  take  part  in  the  changes.  In  fact  each  species  of 
undissociated  molecule  concerned  requires  a  separate  equation 
to  show  its  passage  into,  or  out  of,  the  ionized  condition;  but 
these  equations  may  be  written  together  so  as  to  intersect  and 
give  a  complete  picture  of  the  whole  change. 

In  the  next  section  eight  rules  to  be  observed  in  writing  ionic 
equations  are  given,  and  by  following  these  rules  one  is  able 
to  give,  by  means  of  the  equations  alone,  both  a  fairly  complete 
description,  and  a  remarkably  good  explanation,  of  the  reaction. 

Until  the  student  has  thoroughly  mastered  the  ionic  theory,  he 
should  write  equations  in  the  fully  ionized  intersecting  form  for 
every  reaction  which  he  studies.  Later,  with  the  practice  thus 
acquired,  he  will  be  able  to  interpret  ordinary  single  equations  in 
terms  of  the  ionic  theory. 

Rules  for  Writing  Equations  in  Ionic  Form: 

(1)  Solid  substances  are  underlined:  e.g.,  NaCl. 

(2)  The  un-ionized  part  of  substances  in  solution  is  shown  by 
the  molecular  formula  without  ionic  charges:   e.g.,  HC2H3O2. 

(3)  Ionized  substances  in  solution  are  shown  by  the  formulas 
of  the  ions:  e.g.,  Na+Cl". 

(4)  All  salts,  and  strong  acids  and  bases  (that  is,  those  which 
are  as  much  as  45  percent  ionized  in  0.1  equivalent  solution)  are 
to  be  treated,  as  far  as  equation  writing  is  concerned,  as  if  they 
were    completely   ionized:     e.g.,    Na+Cl~;     H+C1~;     Na+OH~; 
2H+SO4— . 

(5)  In  equations  showing  the  reactions  of  weak  acids  and  bases, 
the  un-ionized  parts  cannot  be  neglected.     Both  the  un-ionized 
and  ionized  parts  must  be  shown,  e.g.: 

HC2H3O2  ^±  H+  +  C2H3O2- 
NH4OH  ^  NH4+  +  OH- 


METATHESIS  87 

(6)  Solid  substances  formed  in  a  reaction  and  thus  precipitated 
are  indicated  by  an  arrow  pointing  downward:  e.g.,  AgCl  J, 

(7)  Gaseous  substances  formed  in  a  reaction  and  thus  escaping 
from  the  solution  are  indicated  by  an  arrow  pointing  upward :  e.g., 

co2t 

(8)  Intersecting  equations.     The  ionization  of  a  substance  is 
itself  a  reaction  which  requires  an  equation;  when  the  products  of 
this  reaction  enter  other  reactions,  the  whole  course  of  the  process 
can  be  shown  only  by  a  series  of  equations.     To  show  their  inter- 
dependence these  equations  should  be  arranged  to  intersect,  the 
formula  of  the  component  common  to  two  equations  being  written 
at  the  point  of  intersection. 

METATHESIS 

Metathesis,  or  double  decomposition,  is  one  of  the  main  types  of 
chemical  reaction,  and  it  takes  place  between  two  compounds,  con- 
sisting merely  in  an  interchange  of  radicals,  the  positive  radical  of 
the  first  compound  pairing  off  with  the  negative  radical  of  the 
second,  and  the  negative  radical  of  the  first  pairing  off  with  the 
positive  of  the  second.  Such  a  reaction  involves  no  change  what- 
ever in  the  valence  of  any  radical  concerned. 

When  solutions  of  two  ionized  substances  are  mixed,  the  oppor- 
tunity is  thereby  furnished  for  the  formation  of  two  new  substances. 
What  will  actually  take  place  depends  on  the  properties  of  these 
two  substances,  as  well  as  upon  the  properties  of  the  original 
substances.  The  various  types  of  metathetical  reactions  are 
classified  on  this  basis. 

Precipitation.  When  a  solution  of  silver  nitrate  is  added  to  a 
solution  of  sodium  chloride  a  voluminous,  curdy,  white  precipitate 
instantly  appears  and  analysis  of  the  precipitate  shows  it  to  be 
silver  chloride. 

The  mixing  together  of  the  solutions  brings  together  the  four 
ions,  Ag+,  NO3~,  Na+,  and  Cl~,  from  which  four  different  pairs,  or 
complete  substances,  are  possible,  AgNO3,  NaCl,  NaNO3,  AgCl, 
Reference  to  a  solubility  table  shows  that  the  first  three  of  these 
are  extremely  soluble,  and  reference  to  the  table  of  ionization  values 
shows  that  each  of  the  three  is  highly  ionized;  on  the  other  hand, 
AgCl  is  highly  insoluble.  The  latter  salt,  therefore,  precipitates 


88  THEORY  OF  IONIZATION 

until  there  are  left  in  solution  only  the  extremely  few  of  its  ions 
corresponding  to  the  solubility  of  silver  chloride. 

Ag+        NO3- 
Cl-        Na+ 

I     f 
AgCli 

For  writing  the  equations  of  ionic  reactions  the  set  of  rules 
given  on  page  86  has  been  devised.  Silver  nitrate  and  sodium 
chloride  are  each  about  86  percent  ionized,  according  to  the  table. 
In  the  equation  these  substances  both  appear  as  completely 
ionized,  the  14  percent  un-ionized  portion  of  each  which  exists 
before  mixing  the  solutions  being  disregarded  (Rule  4).  As  a 
matter  of  fact,  as  soon  as  AgCl  is  precipitated  and  Ag+  and  Cl~ 
ions  are  thus  removed,  the  original  14  percent  of  un-ionized 
AgNO3  and  NaCl  are  no  longer  in  equilibrium  with  the  ions;  fur- 
ther ionization  ensues,  and  the  salts  enter  as  completely  into  re- 
action as  if  they  had  been  100  percent  ionized  at  the  start. 

Sodium  nitrate,  according  to  the  table,  is  86  percent  ionized. 
Hence  14  percent  of  the  total  amount  of  this  salt  must  have  passed 
into  the  un-ionized  condition.  It  will  be  noticed  that  this  fact  is 
neglected  in  the  equation.  For  all  practical  purposes  the  major 
part  of  the  Na+  and  NO3~  ions  are  left  in  exactly  the  same  con- 
dition after  the  reaction  as  they  were  in  the  beginning,  and  it  is 
exactly  this  fact  which  is  emphasized  by  this  manner  of  writing 
the  equation,  for  these  ions  simply  appear  beside  the  oppositely 
charged  ions  with  which  they  were  originally  paired  and  nothing 
happens  to  them.  They  are  placed  adjacent  to  each  other,  show- 
ing that  after  the  changes  have  taken  place  they  are  capable  of 
balancing  each  other  electrically. 

The  arrow  is  indicative  of  the  real  reaction,  and  the  formulas 
of  the  components  which  do  change  are  placed  in  the  line  of  the 
arrow  and  constitute  the  equation  Ag+  -f-  Cl~  — >  AgCl,  which,  for 
convenience,  was  arranged  in  a  vertical  line. 

Experiments  have  shown  that  any  ionized  silver  salt,  e.g., 
Ag2SO4,  AgC2H3O2,  AgClO3,  may  be  substituted  for  AgNO3,  and 
any  ionized  chloride  be  substituted  for  the  NaCl,  and  the  same 
results  will  be  obtained.  The  union  of  Ag+  and  Cl~  ions  to  form 
insoluble  AgCl  is  in  no  wise  affected  by  the  other  ions  with  which 
these  ions  are  at  the  outset  in  electrical  balance.  These  other 


METATHESIS  89 

ions  will  simply  remain  in  the  solution  unless  they,  too,  are  the 
ions  of  some  insoluble  salt,  as,  for  example,  BaSO4. 

The  above  principles  do  not  apply  solely  to  silver  salts  and  chlo- 
rides, but  to  all  solutions.  A  mere  turning  to  a  table  of  solu- 
bilities will  inform  one  whether  a  precipitate  will  be  formed  when 
any  two  solutions  of  electrolytes  are  mixed  together. 

PROBLEMS 

Divide  a  good-sized  sheet  of  paper  into  two  columns;  in  the 
left-hand  column  describe  the  observable  effects  of  bringing 
together  the  substances  noted  in  the  cases  below;  in  the  right-hand 
column  write  the  equation  in  the  fully  ionized  intersecting  form, 
following  Rules  1  to  8,  inclusive,  on  page  86.  In  the  following, 
unless  otherwise  specified,  the  formula  stands  for  the  substance 
in  a  fairly  dilute  solution.  The  Table  of  Solubilities  in  the  Ap- 
pendix should  be  consulted. 

24.  AgNOs  +  NaCl. 

25.  AgCl  (solid)  +  Nal. 

26.  1  mole  Ag(C2H302)  (solid)  +  1  liter  normal  HN03. 

27.  1  mole  Ag2SO4  (solid)  +  2  liters  normal  HNO3. 

28.  BaCl2  +  Na2OO3. 

29.  Pb(N03)2  +  H2S04. 

30.  PbSO4  (solid)  +  Na2S. 

31.  Cd(N03)2  +  NasS. 

32.  CaCl2  +  Na2S04. 

33.  CaSO4  (solid)  +  NasCOg. 

34.  BaSO3  (solid)  +  HC1. 

35.  BaS04  (solid)  +  HC1. 

Neutralization.  Acids  and  bases  have  the  ability  to  mutually 
neutralize  the  distinctive  properties  of  each  other. 

An  acid  tastes  sour,  it  turns  blue  litmus  red,  and  it  imparts  dis- 
tinctive colors  to  a  number  of  other  organic  substances  which  may 
be  used  in  the  same  way  as  litmus;  it  reacts  with  active  metals, 
hydrogen  being  evolved  and  salts  of  the  metal  being  left;  it 
reacts  with  calcium  carbonate  with  an  effervescence  due  to  the 
escape  of  carbon  dioxide.  These  properties  of  an  acid  are  all 
lost  when  the  acid  has  reacted  with  an  equivalent  quantity  of  a 

ise. 

A  base  tastes  alkaline,  that  is,  like  lime  water;    it  turns  red 


90  THEORY  OF  IONIZATION 

litmus  blue,  as  well  as  imparting  distinctive  colors  to  the  other 
organic  substances  which  may  be  used  in  a  similar  manner;  it 
causes  a  slippery  feeling  if  a  drop  of  its  solution  is  rubbed  between 
the  finger  tips.  These  properties  of  a  base  are  all  lost  when  the 
base  has  reacted  with  an  equivalent  quantity  of  an  acid. 

When  the  neutralization  has  been  very  carefully  carried  out,  so 
that  exactly  equivalent  quantities  of  acid  and  base  have  been  used, 
the  resulting  solution  shows  none  of  the  characteristic  properties 
of  either  acid  or  base.  It  still  conducts  electricity  strongly,  show- 
ing that  it  contains  ions;  if  it  is  evaporated  a  solid  salt  is  left. 

Since  we  know  that  all  acids  yield  hydrogen  ions  when  dissolved, 
although  the  negative  ions  may  be  of  most  divergent  kinds,  it  is 
obvious  that  the  distinctive  properties  of  acid  solutions  must  be  the 
properties  of  hydrogen  ions.  Likewise  it  is  obvious  that  the  dis- 
tinctive properties  of  solutions  of  bases  must  be  the  properties  of 
hydroxyl  ions. 

Since  we  may  see  in  the  table  of  ionization  values  that  pure 
water  contains  but  0.000,000,1  mole  of  H~  ions  and  0.000,000,1 
mole  of  OH  ~  ions  per  liter,  we  may  know  in  advance  that  when  an 
acid  and  a  base  are  mixed  the  H+  and  OH~  ions  cannot  remain  in 
the  presence  of  each  other,  but  must  unite  according  to  the  reaction, 


until  only  a  number  corresponding  to  the  exceedingly  small  con- 
centration just  stated  is  left. 

A  salt  may  be  defined  as  a  compound  consisting  of  the  positive 
radical  of  a  base  and  the  negative  radical  of  an  acid.  Hence  the 
products  of  neutralization  are  always  un-ionized  water  and  a  salt. 
But  the  table  of  ionization  values  tells  us  that  all  salts  are  highly 
ionized,  although  acids  and  bases  may  or  may  not  be. 

Neutralization  of  a  Strong  Acid  and  a  Strong  Base.  When  a 
strong,  that  is,  a  highly  ionized  acid  (for  example,  HC1),  is  neutral- 
ized with  a  strong  base  (for  example,  NaOH),  and  the  resulting 
salt,  in  this  case  NaCl,  is  soluble,  the  reaction  consists  essentially 
only  of  the  formation  of  water  from  its  ions. 

H+        Cl- 

OH-        Na+ 

i 
H2O 


METATHESIS 


91 


In  this  connection  a  most  interesting  fact  comes  to  our  attention, 
namely,  that  the  heat  produced  by  the  neutralization  of  one 
equivalent  of  any  strong  base  with  one  equivalent  of  any  strong 
acid  is  always  the  same,  namely,  13,700  calories.  That  the  heat 
effect  is  the  same  is  in  itself  a  strong  indication  that  the  reaction 
is  in  each  case  the  same,  and  this  fact,  then,  is  in  entire  accord  with 
our  conception  of  the  reaction  of  neutralization.  In  the  follow- 
ing table  are  given  some  of  the  measured  values  of  the  heat  of 
neutralization  of  acids  and  bases,  both  weak  and  strong. 

Heat  evolved  by  the  neutralization  of  one  equivalent  of 
acid  with  one  equivalent  of  base  (in  calories) 


HC1 

HNO3 

HC2H302 

H2S 

NaOH.  . 

13,700 

13,700 

13  300 

3  800 

KOH  

13,700 

13,800 

13,300 

3  800 

NH4OH 

12,400 

12500 

12  000 

3  100 

Neutralization  of  a  Weak  Acid  or  a  Weak  Base.  Weak  acids 
and  weak  bases  are  but  sparingly  ionized.  Acetic  acid  is  typical 
of  a  rather  weak  acid,  being  1.4  percent  ionized  in  0.1  normal 
solution.  Ammonium  hydroxide  is  typical  of  a  rather  weak  base, 
it  having  the  same  degree  of  ionization,  namely,  1.4  percent  in  a 
0.1  normal  solution,  as  acetic  acid.  Neither  acetic  acid  nor  am- 
monium hydroxide  solution  conducts  the  current  strongly,  but 
if  the  two  solutions  are  mixed,  we  observe  what  is  a  rather  startling 
fact  if  we  have  not  thought  out  in  advance  what  to  expect,  namely, 
that  the  resulting  solution  is  a  strong  conductor. 

Neither  the  acid  nor  the  base  alone  furnishes  many  ions,  but 
when  they  are  mixed,  the  H+  and  OH~  ions  present  unite  at  once. 
This  leaves  the  un-ionized  parts  of  the  acid  and  base  out  of  equi- 
librium and  further  ionization  occurs  in  consequence.  This 
process  continues  until  both  the  acid  and  the  base  have  become 
fully  ionized  because  there  can  be  no  accumulation  of  H+  and  OH~ 
ions  in  presence  of  each  other. 

The  neutralization  of  a  weak  acid  and  a  weak  base  then  con- 
sists of  three  simultaneous  but  distinct  ionic  reactions:  the  ion- 
ization of  the  acid,  the  ionization  of  the  base,  and  the  formation 
of  water  from  its  ions.  These  reactions  may  be  arranged  in  an 


92  THEORY  OF  IONIZATION 

intersecting  form  in  order  to  show  which  of  the  components  take 
part  simultaneously  in  two  reactions: 

HAc          —    H+       +     Ac- 
NH4OH    ;=>    OH-     +     NH4+ 

|  H20 

As  this  train  of  reactions  proceeds  it  is  obvious  that  NH4+  and 
Ac~  ions,  that  is,  the  ions  of  the  salt  ammonium  acetate,  ac- 
cumulate in  the  solution,  and  that  their  presence  accounts  for  the 
high  conductivity  of  the  neutral  solution. 

The  total  heat  effect,  12,000  calories,  produced  by  the  action  of 
one  equivalent  of  acetic  acid  and  one  equivalent  of  ammonium 
hydroxide,  is  the  sum  of  the  heat  effects  of  the  three  separate  reac- 
tions, and  we  should  expect  the  value  to  be  different  from  the 
value  for  the  neutralization  of  a  strong  acid  and  a  strong  base. 
Since  13,700  calories  must  be  generated  by  the  formation  of  1 
mole  of  water,  the  difference  between  this  value  and  12,000,  or  1700 
calories  must  have  been  absorbed  in  the  ionizing  of  the  acid  and 
the  base. 

Displacement  of  a  Weak  Acid  from  its  Salt.  Sodium  acetate  is 
a  salt  of  acetic  acid,  but  a  solution  of  this  salt  has  none  of  the 
pungent  odor  of  acetic  acid,  not  even  if  it  is  warmed.  If,  however, 
a  strong  acid  like  sulphuric  acid  is  added  to  the  solution,  an  unmis- 
takable odor  of  acetic  acid  is  at  once  manifest. 

The  neutral  salt  contains  acetate  ions,  but  the  attraction  of  op- 
positely charged  sodium  ions  prevents  them  from  escaping  from 
the  solution  to  give  an  odor,  even  if  one  could  expect  charged  ions 
to  have  any  of  the  same  properties  as  uncharged  HAc  molecules. 
With  the  addition  of  sulphuric  acid,  however,  the  ions  of  acetic 
acid  are  brought  into  the  presence  of  each  other,  and  they  must 
at  once  reach  a  condition  of  equilibrium  with  the  un-ionized  mole- 
cules. From  the  table  of  ionization  values  we  see,  therefore, 
that  100  —  1.4  =  98.6  percent  of  these  ions  unite  to  form  un- 
ionized acetic  acid ;  and  since  this  substance  is  known  to  be  some- 
what volatile  and  to  possess  a  very  powerful  odor,  the  observed 
effect  is  accounted  for. 

2H+         SO4— 
2Ac~        2Na+ 

0 

2HAc 


METATHESIS  93 

This  process  is  usually  spoken  of  as  the  displacement  of  the  weak 
acid  from  its  salt;  the  ions  of  the  salt  of  the  stronger  acid  remain 
in  the  solution.  If  we  were  always  to  think  solely  from  the  ionic 
viewpoint,  perhaps  we  should  want  to  describe  this  process  as  the 
formation  of  the  weak  acid  from  its  ions. 

Displacement  of  a  Weak  Base  from  its  Salt.  An  example  of  this 
kind  of  process  is  given  by  the  action  of  the  strong  base,  sodium 
hydroxide,  upon  a  solution  of  the  salt,  ammonium  chloride.  The 
salt  solution  is  odorless,  but  the  odor  of  ammonia  is  observed  as 
soon  as  the  strong  base  is  added. 

NH4+  Cl- 

OH-  Na+ 

Jt 

NH4OH 

In  this  case,  as  in  the  case  of  the  weak  acid,  the  major  part  of  the 
ions  of  the  weak  base  combine  to  form  un-ionized  molecules. 

The  strong  odor  is  due  to  the  escape  of  small  amounts  of  am- 
monia, NH3;  from  the  solution.  Ammonium  hydroxide  is  capable 
of  undergoing  two  kinds  of  dissociation:  the  electrolytic  disso- 
ciation, or  ionization,  which  we  have  already  discussed,  and  a 
non-electrolytic  dissociation, 

NH4OH  <=>  H20  +  NH3. 

The  latter  sort  of  dissociation  is  subject  to  the  same  rules  of  equi- 
librium as  is  ionization,  and  we  can  have  u-n-ionized  ammonium 
hydroxide  at  the  same  time  in  equilibrium  with  two  sets  of  dis- 
sociation products, 

NH4+  +  OH-  +±  NH4OH  <±  NH3  +  H20. 

Therefore,  whenever  NH4+  and  OH~  ions  are  brought  together, 
they  must  come  to  equilibrium  with  a  large  proportion  of  NH4OH 
and  the  latter  must  come  to  equilibrium  with  NH3  and  H2O  and, 
through  the  NH4OH,  the  NH3  and  H20  must  be  in  equilibrium 
with  the  ions.  In  dilute  solutions  where  the  proportion  of  H2O 
is  large,  the  amount  of  NH3  necessary  to  produce  equilibrium  is 
small.  Such  a  substance  as  pure  ammonium  hydroxide  of  the 
composition  shown  by  the  formula  is  unknown,  because,  if  it 
existed  for  a  moment,  it  would  at  once  undergo  non-electrolytic 
dissociation  until  it  came  to  a  state  of  equilibrium  with  the  prod- 


94  THEORY  OF  IONIZATION 

ucts  H2O  and  NH3,  the  first  product  remaining  as  a  part  of  the 
NH4OH  solution,  and  the  larger  part  of  the  NH3  escaping  as  gas. 
The  more  water  present,  however,  obviously  the  less  NH3  is 
necessary  to  maintain  equilibrium  with  the  NH4OH. 

Basic  Properties-  of  Metal  hydroxides.  The  metallic  elements 
as  a  class  are  characterized  chemically  in  that  their  hydroxides  are 
basic.  The  hydroxides  of  the  alkali  metals,  of  which  sodium  and 
potassium  are  the  most  common,  are  very  sjbrong  and  very  soluble 
bases.  The  hydroxides  of  the  alkaline  earth  elements,  calcium, 
strontium,  and  barium,  are  very  strong  bases,  but  they  are  not  as 
soluble  as  the  alkali  metal  hydroxides. 

The  hydroxides  of  the  heavy  metals  are,  as  a  rule,  very  insoluble 
and  very  weakly  basic.  Addition  of  a  soluble  base,  even  as  weak 
a  base  as  ammonium  hydroxide,  may  be  expected  to  displace  the 
hydroxide  of  a  heavy  metal  from  a  solution  of  its  salt.  Thus 
when  ammonium  hydroxide  is  added  to  a  solution  of  ferric  chlo- 
ride a  voluminous  reddish  brown  precipitate  of  ferric  hydroxide 
separates. 

Fe+++     3C1- 
3NH4OH  ^±  3OH-     3NH4+ 

I 
Fe(OH)3  I 

If  an  amount  of  ammonium  hydroxide  equivalent  to  the  ferric 
chloride  is  taken,  this  reaction  runs  so  nearly  to  completion  that  no 
detectable  trace  of  Fe+++  ions  is  left  in  the  solution. 

The  hydroxides  of  such  metals  as  iron,  copper,  silver,  and  lead 
do  not  show  to  any  marked  degree  the  characteristic  properties  of 
turning  litmus  blue,  or  of  alkaline  taste,  or  of  producing  a  slippery 
feeling  when  rubbed  between  the  finger  tips.  Their  main  claim 
to  classification  as  bases  lies  in  their  ability  to  neutralize  acids;  to 
illustrate,  let  us  consider  the  cases  of  ferric  hydroxide  and 
copper  oxide. 

If  the  ferric  hydroxide  produced  by  the  addition  of  ammonium 
hydroxide  to  a  ferric  chloride  solution  is  collected  on  a  filter  and 
washed  with  water,  a  reasonably  pure  substance  of  the  composition 
given  by  the  formula  Fe(OH)3  is  obtained.  If  this  substance  is 
rinsed  into  a  beaker  with  a  considerable  amount  of  pure  water  and 
stirred,  a  thick  reddish  brown  suspension  is  obtained,  but  after  this 
has  stood  for  some  time  it  separates,  the  insoluble  Fe(OH)3  col- 


METATHESIS  95 

lecting  at  the  bottom,  clear  water  only  remaining  in  the  upper  part 
of  the  beaker.  Litmus  dipped  into  the  liquid  is  unaffected.  If 
now  such  an  amount  of  hydrochloric  acid  is  added  to  the  beaker 
that  3  moles  of  HC1  are  used  for  every  mole  of  Fe(OH)3,  and  the 
whole  is  stirred,  the  reddish  brown  substance  dissolves  and  a 
perfectly  clear  yellow  solution  fills  the  beaker.  Furthermore,  this 
solution  is  not  strongly  acid,  as  we  should  expect  had  the  Fe(OH)3 
not  been  suspended  in  the  water  to  which  the  HC1  was  added. 
It  is  true  that  this  solution  will  color  litmus  red,  showing  that  the 
acid  is  not  fully  neutralized,  but  an  accurate  measurement  of  the 
H+  ion  concentration  (by  methods  which  we  will  not  elaborate 
here)  will  show  that  it  corresponds  not  to  the  strength  of  hydro- 
chloric acid,  but  very  nearly  to  that  of  the  weak  acetic  acid.  The 
very  weak  base,  Fe(OH)3,  then  is  able  to  neutralize  to  a  large 
extent  the  effects  of  a  strong  acid.  Why  the  acid  is  not  completely 
neutralized  will  appear  under  the  discussion  of  hydrolysis  in  the 
next  section. 

Fe(OH)3  <=±  Fe+++    +    30H~ 
3C1-       +    3H+ 

ir 

3H20 

If  black,  finely  powdered  copper  oxide,  CuO,  is  stirred  with 
water,  it  settles  out  again  very  quickly,  and  the  water  has  not 
acquired  sufficient  basic  properties  to  affect  litmus.  If  an  equiva- 
lent amount  of  hydrochloric  acid,  that  is,  2  moles  for  every  mole 
of  CuO,  is  added  and  thoroughly  stirred,  the  solution  immediately 
acquires  a  blue  green  color  and  in  sufficient  time  the  whole  of  the 
black  copper  oxide  is  dissolved.  This  solution,  like  the  one 
previously  discussed,  shows  very  little  acid  strength. 

Copper  oxide  is  not  itself  a  base,  but  it  is  very  closely  allied  to 
copper  hydroxide, 

CuO  +  H20  ^±  Cu(OH)2. 

Mere  contact  with  water  does  not  seem  to  cause  the  copper  oxide 
to  change  appreciably  to  the  hydroxide,  but  we  can  at  least  assume 
a  slight  tendency  for  this  to  happen.  Addition  of  acid  would 
neutralize  the  copper  hydroxide  and  removal  of  the  latter  would 
allow  the  conversion  of  more  copper  oxide  to  hydroxide;  thus 
the  train  of  reactions  may  be  pictured  as  follows: 


96  THEORY  OF  IONIZATION 

CuO  +  H20  ^±  Cu(OH)2  ^±  Cu++  +  20H- 

2C1-       2H+ 

jr 

2H2O 

If  the  solution  of  copper  salt  is  kept  at  ordinary  temperature  the 
addition  of  a  soluble  base  causes  a  light  blue  precipitate  to  appear, 
having  a  variable  composition  but  approximating  to  the  formula 
[Cu(OH)2]3.CuCl2.  If  we  direct  our  attention  particularly  to  the 
Cu(OH)2  in  the  formula,  the  reaction  becomes  essentially  a  com- 
bination of  Cu++  and  OH~  ions  to  form  the  weak  base  Cu(OH)2. 

The  oxides  of  the  alkali  and  alkaline  earth  metals  combine  ener- 
getically with  water  to  form  hydroxides  which  are  strong  bases. 
By  comparison  the  oxides  of  the  heavy  metals  combine  sluggishly 
with  water,  but  the  oxides  themselves  may  be  said  to  possess 
somewhat  the  same  basic  properties  as  the  hydroxides,  because 
they  do  have  the  property  of  neutralizing  acids. 

Formation  of  Volatile  Products.  When  a  product  of  a  reaction 
is  volatile  it  has  a  tendency  to  escape  from  the  sphere  of  action, 
and  the  progress  of  the  reaction  towards  the  formation  of  this 
product  is  favored,  as  in  the  case  of  precipitation. 

When  the  products  of  a  reaction  taking  place  in  solution  are  all 
soluble,  their  concentration  will  increase  as  the  reaction  pro- 
gresses, until  a  point  of  equilibrium  is  reached,  at  which  point  the 
products  react  with  each  other  to  form  the  original  substances 
again,  and  the  backward  reaction  takes  place  with  sufficient 
rapidity  just  to  offset  the  effect  of  the  forward  reaction. 

But  with  the  complete  escape  of  one  of  the  products  as  a  gas, 
just  as  in  the  case  of  precipitation,  and  consequent  removal  of 
one  of  the  products  from  the  sphere  of  action,  the  reverse  reaction 
is  eliminated,  and  the  forward  reaction  is  thus  enabled  to  run  to 
completion. 

PROBLEMS 

Describe  the  observable  effects  and  write  the  fully  ionized  equa- 
tions for  the  following  cases: 

36.  HN03  +  NaOH. 

37.  HC2H3O2  +  NH4OH. 

38.  Mg(OH)2  (solid)  +  HN03. 

39.  CuO  (solid)  +  H2SO4. 


COMPLEX  IONS  97 

40.  H2S  (gas)  +  NaOH. 

41.  NH3  (gas)  +  HC1. 

42.  H(C7H5O2)  (solid)  +  NaOH. 

Benzole  acid  H(C7H502)   is  sparingly  soluble  and  a  little 
stronger  than  acetic  acid. 

43.  K(C7H502)  +  HC1. 

44.  Ca(C2H302)2  +  HNO3.     Note  odor. 

45.  (NH4)2  SO4  +  NaOH.      Note  odor. 

46.  NH4C1  (in  excess)  +  Ca(OH)2  (solid). 

47.  CaCO3  (solid)  +  HN03. 

48.  CaC03  (solid)  +  HC2H302. 

49.  FeS  (solid)  +  HC1. 

50.  MgCl2  +  NaOH. 

51.  MgCl2  +  NH4OH. 

52.  MgCl2  +  NH4C1  +  NH4OH. 

53.  Ca(OH)2  (solid)  +  FeCl3. 

54.  Ca(OH2)  (saturated  solution)  +  CO2  (gas,  in  moderate 
amount  and  then  in  excess). 


COMPLEX  IONS 

Ammoniates.  Review  what  was  said  in  Chapter  II  about 
water  of  crystallization  and  hydrates.  Two  definite  crystalline 
compounds  of  sodium  carbonate  and  water  are  the  monohydrate 
and  the  decahydrate.  Both  of  these  will  dissolve  in  water  but 
both  solutions  are  absolutely  identical.  Furthermore,  either  one 
or  the  other  of  these  hydrates  can  be  caused  to  crystallize  from 
the  same  solution  by  adjusting  the  temperature. 

In  solution,  undoubtedly  some  water  is  in  combination  with 
the  salt,  but  it  is  impossible  to  say  how  much,  because  there  is  no 
physical  means  of  distinguishing  the  water  thus  held  in  com- 
bination from  the  solvent  water. 

There  are  other  substances  than  water,  notably  ammonia, 
which  form  compounds  similar  to  hydrates,  in  this  case  ammo- 
mates.  Numerous  solid  compounds  containing  ammonia  of 
crystallization  are  known,  for  example  ammonio-copper  sulphate 
CuS04.4NH3.H20  which  is  the  subject  of  one  of  our  later  prepa- 
rations. This  substance  is  easily  soluble  in  water  containing  a 
little  excess  of  ammonia,  and  it  is  possible  to  measure  how  much 
ammonia  is  held  to  the  salt,  or  rather  to  the  positive  ion,  in  the 


98  THEORY  OF  IONIZATION 

solution.  This  may  be  done,  for  example,  by  passing  an  electric 
current  through  the  solution  and  measuring  the  proportionate 
amounts  of  copper  and  ammonia  which  travel  with  the  positive 
current.  Results  show  that  4  moles  of  NH3  travel  with  each  mole 
of  Cu++  ions.  The  formula  of  the  deep  blue  ammonio-copper 
ion  is  therefore  Cu.4NH3++. .' 

When  a  moderate  amount  of  ammonia  is  added  to  a  copper  salt 
solution,  a  light  blue  precipitate  is  formed  which  is  really  a  basic 
salt  but  for  simplicity  we  shall  treat  as  the'-simple  hydroxide. 

Cu++          SO4- 
2NH4OH^±20H-         2NH4+ 

Cu(OH)2 1 

Addition  of  more  ammonia  quickly  causes  this  precipitate  to 
redissolve  giving  an  intensely  deep  blue  solution 

Cu(OH)2  ^  Cu++  +  20H- 
4NH4OH  ^  4NH3°  +  4H2O 

If 

(Cu.4NH3)++ 

The  hydroxide  of  the  ammonio-cupric  radical,  unlike  that  of  the 
simple  cupric  radical,  is  very  soluble  and  highly  ionized. 

If  the  two  sets  of  ionic  equations  are  now  put  together  it  is 
seen  that  2NH4+  and  20H~  have  been  left  and  these  will  combine 
to  form  2NH4OH,  which  will  cancel.  The  total  result  when  four 
moles  of  ammonia  are  added  to  one  mole  of  copper  salt  is  given 
by  the  equation 

Cu++      SO4— 
4NH4OH  ^±  4NH3  +  4H2O 

I 
(Cu.4NH3)++ 

the  resulting  solution  containing  ionized  (Cu.4NH3)S04  and  only 
enough  excess  of  ammonia  to  prevent  the  dissociation  of  this 
ammoniate. 

It  is  interesting  to  note  that  if  freshly  precipitated  and  thor- 
oughly washed  Cu(OH)2  is  treated  with  ammonia,  a  similar  deep 
blue  solution  is  obtained,  but  in  this  case  the  negative  ions  are 
hydroxyl  instead  of  sulphate.  Ammonio-copper  hydroxide  is 
very  soluble  and  very  highly  ionized,  and  the  solution  compares 


COMPLEX  IONS  99 

in  basic  strength  with  one  of  sodium  hydroxide.  Such  a  solution 
is  known  as  Schweitzer's  reagent  and  it  has  the  remarkable  prop- 
erty of  being  able  to  dissolve  cellulose  (wood  fiber  and  filter  paper 
are  composed  of  cellulose). 

It  is  most  interesting  that  the  addition  of  ammonia  to  the  simple 
ions  of  several  of  the  heavy  metals  produces  similar  effects.  The 
basic  character  of  the  metals  seems  thereby  to  be  greatly  strength- 
ened. The  following  list  gives  the  metals  which  possess  this 
property  to  a  marked  degree  and  also  the  formulas  of  their  am- 
monio  ions : 


copper 

(ous) 

(Cu.2NH3)  + 

(ic) 

(Cu.4NH3)++ 

silver 

(Ag.2NH3)  + 

zinc 

(Zn.4NH3)++ 

cadmium 

(Cd.4NH3)++ 

nickel 

(Ni.4NH3)++ 

cobalt 

(ous) 

(Co.4NH3)++ 

(ic)* 

(Co.6NH3)+++ 

Aluminum,  iron,  tin,  lead,  and  some  other  metals  seem  to  be 
wholly  devoid  of  the  power  of  forming  such  ammonio  compounds. 

Complex  Negative  Ions.  When  an  equivalent  amount  of 
potassium  cyanide  is  added  to  a  solution  of  silver  nitrate,  quan- 
titative precipitation  of  silver  cyanide  takes  place 

Ag+        N03- 
CN-      K+ 
I 

AgCNj 

When  another  equivalent  of  potassium  cyanide  is  added  the 
precipitate  entirely  redissolves.  In  this  solution  the  silver  is 
found  to  be  in  the  negative  ion,  and  to  be  associated  with  2  cya- 
nide radicals. 

AgCN  ?±  AgCN 

CN-       K+ 

I 

(Ag(CN)2)- 

We  assume  that  the  very  few  dissolved  un-ioriized  AgCN  molecules 
in  equilibrium  with  the  solid,  combine  with  CN~  ions  to  form  the 

*  The  ammonio-cobaltic  and  similar  cobaltic  compounds  are  very  nu- 
merous and  very  complicated;  the  formula  given  represents  the  most  typical. 


100  THEORY  OF  IONIZATION 

complex  ion  (Ag(CN)2)~.  Continual  removal  of  AgCN  mole- 
cules from  the  solution  prevents  an  equilibrium  with  the  solid 
being  restored  and  the  latter  thus  dissolves  completely. 

This  example  is  typical  of  complex  ion  formation.  The  for- 
mulas of  a  few  complex  ions,  in  the  decreasing  order  of  their 
stability  are: 

Co(CN)«- 

Fe(CN)6 ,. 

Fe(CN)6 

Ag(CN)2- 

Cu(CN)2- 

Ni(CN)4— 

AgI2~ 

AgCl2- 

The  last  in  the  list  is  so  little  stable  that  it  can  only  exist  in 
a  concentrated  chloride  solution.  Dilution  of  such  a  solution 
causes  precipitation  of  all  the  silver  as  simple  silver  chloride. 

The  ions  NO3~,  SO4  ,  P04  are  so  very  stable  that  we  do 
not  think  of  them  as  other  than  simple  ions.  Nevertheless  they 
are  obviously  complex  and  they  are  probably  similar  in  nature  to 
the  complex  ions  just  considered  except  for  the  fact  that  any  simple 
ions  out  of  which  they  might  be  supposed  to  be  built  have  never 
been  identified. 

PROBLEMS 

Describe  observable  effects  and  write  the  fully  ionized  equa- 
tion for  the  following  cases,  when  the  reagent  (second  formula) 
is  added  (a)  in  limited  amount,  and  (6)  in  excess. 

55.  ZnS04  +  NH4OH. 

56.  CuCl  solid  +  NH4OH. 

57.  Ag2SO4  +  NH4OH. 

58.  (Cu.4NH3)SO4  +  HNO3. 

59.  (Ag.2NH3)Cl  +  HN03. 

60.  AgNO3  +  KI. 

61.  AgCl  (solid)  +  NaCl  (saturated  solution,  in  very  large 
excess). 

62.  FeSO4  +  KCN. 

63.  FeCls  +  KCN. 


HYDROLYSIS  101 

HYDKOLYSIS 

The  ionization  of  water  is  so  small  that  in  most  cases  it  can  be 
totally  disregarded.  It  cannot  be  neglected,  however,  in  solu- 
tions of  salts  when  either  the  acid  or  the  base  —  or  both  —  from 
which  the  salt  is  derived,  is  extremely  weak. 

Potassium  cyanide  is  the  salt  of  the  weak  hydrocyanic  acid, 
HCN  (ionization  =  0.01  percent  in  0.1  equivalent  solution), 
and  the  strong  base,  potassium  hydroxide.  A  solution  of  this 
salt  shows  an  alkaline  reaction  to  litmus,  thus  demonstrating 
that  the  solution  contains  an  appreciable  quantity  of  OH~  ions. 
This  is  the  result  of  hydrolysis,  and  the  process  may  be  explained 
as  follows: 

K+        CN- 
H20^±OH-         H+ 

IT 

HCN 

The  salt,  in  accordance  with  the  general  rule  for  salts,  will  exist 
in  solution  in  the  ionized  condition.  Water  is  in  equilibrium  with 
an  almost  infinitesimal  number  of  its  own  ions.  But  even  the  small 
number  of  H+  ions  thus  furnished  to  the  solution  is  more  than  can 
exist  in  presence  of  the  large  concentration  of  CN~  ions  of  the 
salt.  Undissociated  hydrocyanic  acid,  HCN,  must  form;  but 
since  this  removes  some  of  the  H+  ions,  the  equilibrium  between 
water  and  its  ions  is  temporarily  destroyed.  The  equilibrium 
must  be  immediately  reestablished  through  the  ionization  of  more 
water.  This  cycle  of  reactions  repeats  itself  a  great  many  tunes 
until  complete  equilibrium  among  all  the  components  is  estab- 
lished. When  this  condition  is  reached,  as  really  happens  in  a 
very  short  time,  there  has  been  a  considerable  accumulation  of 
OH~  ions  and  of  an  equivalent  amount  of  un-ionized  HCN. 

In  order  to  fully  comprehend  the  extent  and  the  limitation  of 
this  hydrolysis,  we  should  consider  the  reverse  reaction  which 
occurs  when  solutions  of  hydrocyanic  acid  and  potassium  hydrox- 
ide are  mixed: 

K+        OH- 
HCN    ^±    CN-        H+ 

i 

H20 


102  THEORY  OF   IONIZATION 

The  few  ions  furnished  by  the  acid  combine  at  once  with  OH~ 
ions  of  the  base  to  form  water,  and  this  removal  of  H  +  ions  allows 
more  of  the  acid  to  ionize.  This  cycle  of  operations  repeats  itself 
until  we  have  the  same  state  of  equilibrium  as  existed  in  the  solu- 
tion obtained  by  dissolving  pure  potassium  cyanide  in  pure  water. 

The  reaction  of  neutralization  in  this  case  goes  about  99  percent 
of  the  way  to  completion  when  equivalent  amounts  of  the  acid  and 
base  are  mixed.  The  reverse  reaction,  that  is,  the  hydrolysis  of 
potassium  cyanide,  progresses  only  about  1  percent  of  the  way  to 
completion  before  the  state  of  equilibrium  is  reached. 

If  we  consider  the  case  of  a  salt  of  a  much  weaker  acid  than 
hydrocyanic  acid,  or  of  a  salt  of  both  a  very  weak  acid  and  a  very 
weak  base,  it  is  fairly  obvious  that  hydrolysis  will  be  much  more 
extensive. 

Aluminum  sulphide  furnishes  a  good  example  of  this,  for  when 
it  is  treated  with  water  its  hydrolysis  is  complete  : 


A12S3  ^  A12S3  ^±  2A1+++  3S- 

6H20^±60H-  6H+ 

I  I 

2A1(OH)3  3H2S 

I  I 

2A1(OH)3  1          3H2S  t 

Note  in  this  reaction  that  solid  A12S3  disappears,  and  solid  Al(OH)3i 
and  gaseous  H2S  j  appear. 

IONIZATION  OF  POLYBASIC  ACIDS 

It  is  usually  true  with  polybasic  acids  that  one  hydrogen  radical 
ionizes  with  greater  facility  than  the  remaining  ones.  Thus 
phosphoric  acid  ionizes  primarily  as  a  monobasic  acid 

H3PO4;=±H++/H2PO4- 

to  the  extent  of  27  percent  in  0..1  equivalent  solution.  The  fairly 
high  concentration  of  H+  ions  thus  established  prevents  appre- 
ciable ionization  of  the  H2PO4~  ion,  but  if  one  equivalent  of  NaOH 
is  added  for  each  mole  of  H3P04,  the  H+  ions  from  the  first  H 
radical  are  entirely  removed  and  the  H2PO4~  ion  itself  ionizes 


IONIZATION  OF  POLYBASIC  ACIDS  103 

to  the  extent  of  0.2  percent.  If  a  second  equivalent  of  NaOH 
is  now  added,  the  HPO4  ion  is  enabled  to  ionize  to  the  extent 
of  0.0002  percent. 

When  the  hydrolysis  of  the  salt  of  a  polybasic  acid  is  considered, 
the  different  H  radicals  must  be  treated  separately.  Thus  when  1 
mole  of  tertiary  sodium  phosphate,  Na3PO4,  is  dissolved  in  water, 
hydrolysis  takes  place  very  extensively  as  follows: 

Na+      Na+      Na+     PO4— 
H20^±OH-  H+ 

\\ 
HP04— 

The  solution  will  have  a  very  strong  alkaline  reaction,  since  it 
contains  a  large  fraction  of  1  mole  each  of  ionized  NaOH  and  of 
ionized  secondary  sodium  phosphate,  Na2HP04.  The  OH~  ions 
thus  formed  check  the  hydrolysis  of  the  secondary  sodium  phos- 
phate; but  if  solid  secondary  sodium  phosphate  is  dissolved  in 
water,  hydrolysis  of  this  salt  ensues  to  a  sufficient  extent  to  make 
the  solution  alkaline  to  litmus. 

Na+         Na+         HPO4~ 
H20    ^    OH-  H+ 

\\ 
H2P04- 

When  primary  sodium  phosphate,  NaH2PO4,  is  dissolved,  a 
weakly  acid  solution  is  obtained,  this  effect  being  due  to  the 
tendency  of  the  second  hydrogen  radical  of  the  acid  to  ionize. 

Na+         H2PO4- 

ir 

H+ 
HP04— 

PKOBLEMS 

When  the  following  salts  are  dissolved  in  water  >  decide  from  a 
consideration  of  the  degree  of  ionization  of  the  base  and  acid  con- 
cerned in  each  case  whether  the  solution  will  be  neutral,  weakly 
acidic,  strongly  acidic,  weakly  basic,  or  strongly  basic,  and  give 
the  explanation  with  a  fully  ionized  equation. 

64.  KN03.  66.   NH4C2H302. 

65.  Ca(CN)2.  67.  A1C13. 


104  THEORY  OF  IONIZATION 

68.  A1(C2H3O2)3.  73.   NaH2PO4. 

69.  Na3AsO4.  74.   Na2HPO4. 

70.  Na^COs.  75.   Na3P04. 

71.  NaHCO3.  76.   AgNO3. 

72.  NaHS04.  .A 

REACTIONS  OF  OXIDATION  AND  REDUCTION 
Ionic  Displacement.  Electromotive  Series.  When  a  strip  of 
zinc  is  placed  in  a  solution  of  copper  sulphate,  it  is  noticed  that 
a  spongy  deposit  of  copper  metal  soon  appears  on  the  surface  of 
the  zinc,  and  that  the  solution  loses  its  blue  color.  Then  if  the 
solution  is  tested  for  the  presence  of  copper  and  zinc  ions  by  add- 
ing hydrogen  sulphide,  it  is  found  that  this  reagent  gives  a  white 
precipitate.  This  test  shows  that  copper  ions  are  now  absent 
and  that  zinc  ions  are  present  because  we  know  hydrogen  sul- 
phide will  precipitate  black  copper  sulphide  from  a  solution  of 
copper  ions,  and  white  zinc  sulphide  from  a  solution  of  zinc  ions. 
Since  ordinary  pieces  of  metal  are  not  charged,  it  is  obvious  that 
the  reaction  has  consisted  in  a  transfer  of  the  positive  charges  of 
the  copper  ions  to  the  zinc  atoms,  or,  more  strictly,  of  negative 
electrons  from  the  zinc  atoms  to  the  copper  ions: 

Cu++  S04- 

Zn° 

I 

Cu° 

Zn++ 

The  small  zero  mark  °  is  not  essential;  it  may  be  used  when  it  is 
desired  to  attract  particular  attention  to  the  fact  that  the  atom  to 
which  it  is  attached  is  not  electrically  charged.  Since  the  S04~~ 
ions  take  no  part  in  the  above  reaction  beyond  balancing  by  their 
charges  the  positive  charges  of  the  metal  ions,  we  can  eliminate 
them  from  the  equation,  which  then  becomes  simplified  to : 

Cu++  +  Zn_°  -»  Cu°  |  +  Zn++. 

The  metals,  including  hydrogen,  may  be  arranged  in  the  order  in 
which  they  strive  to  pass  into  the  ionic  condition,  as,  in  this  case, 
zinc  does  at  the  expense  of  copper.  Such  a  series  is  known  as  the 
Electromotive  Series,  because  the  electromotive  force  of  such 
reactions,  if  properly  disposed  in  a  cell,  may  be  made  to  send  a 
current  through  an  external  wire  connector.  (See  Electromotive 
Series,  Appendix  p.  310.) 


OXIDATION  AND  REDUCTION  105 

A  characteristic  of  metallic  elements  is  that  they  can  form  simple 
positive  ions,  but  never  simple  negative  ions.  In  other  words,  a 
metal  atom  may  lose  one  or  more  negative  electrons,  but  it  can 
never  attach  to  itself  electrons  in  excess  of  those  forming  the  make- 
up of  the  unelectrified  atom. 

On  the  other  hand,  a  characteristic  of  some  of  the  most  pro- 
nouncedly non-metallic  elements,  viz.  :  F,  Cl,  Br,  I,  S,  is  that 
they  can  form  simple  negative  ions.  No  non-metal  ever  forms 
simple  positive  ions.  The  non-metals  may  be  arranged  in  a 
negative  electromotive  series. 

The  broadest,  as  well  as  the  simplest,  definition  of  oxidation  is 
the  increasing  of  the  positive  valence  of  an  element.  This  can  be 
accomplished  only  through  the  simultaneous  and  equivalent  de- 
crease of  the  valence  of  another  element  involved  in  the  reaction. 
The  other  element  is  said  to  be  reduced.  Thus  oxidation  and 
reduction  always  occur  together;  one  cannot  occur  alone.  In  the 
above  definition,  positive  valence  is  considered  from  an  algebraic 
standpoint.  Thus  iodine  in  an  iodide  is  said  to  be  oxidized  when 
its  valence  is  changed  from  minus  one  to  zero;  as,  for  example,  in 
the  reaction, 

2K+ 


If  we  take  the  viewpoint  that  valence  is  due  to  the  attraction  of 
electrical  charges  on  the  atoms,  the  difference  between  reactions  of 
oxidation  and  reduction  and  of  metathesis  resolves  itself  into  this  : 
oxidation  and  reduction  involve  a  transfer  of  charges  from  one  atom 
to  another;  metathesis  involves  no  transfer  of  charges,  but  simply  a 
regrouping  of  the  charged  radicals. 

The  simplest  type  of  oxidation  and  reduction  reaction  is  that 
which  involves  merely  the  charging  and  discharging  of  simple 
ions.  The  course  of  such  a  reaction  can  be  predicted  from  a 
knowledge  of  where  the  elements  concerned  stand  in  the  electro- 
motive series. 

PEOBLEMS 

In  the  left-hand  column  state  the  observable  effect  and  in  the 
right-hand  column  write  the  fully  ionized  equations. 

77.  Fe  (metal)  +  CuS04. 

78.  Zn  (metal)  +  HC1. 


106  THEORY  OF  IONIZATION 

79.  Cu  (metal)  +  HC1. 

80.  Ag  (metal)  +  AuCl3. 

81.  Cu  (metal)  +  PtCl4. 

82.  AgCl  (finely  divided  solid)  +  Zn  (zinc  dust). 

83.  Ag  (in  a  photograptdc  print)  +  PtCU. 

84.  FeCl2  +  C12. 

85.  CuCl  (solid)  +  C12. 

86.  Na  (metal)  +  H2O 

LAW  OF  MOLECULAR  CONCENTRATION 

The  treatment  of  equilibrium  in  the  preceding  pages  has  been 
qualitative,  having  been  based  on  the  obvious  principle  that  the 
tendency  of  a  reversible  reaction,  say 

A+B^C+D 

to  proceed  to  the  right  is  determined  by  the  concentrations  of  A 
and  B,  whereas  the  tendency  to  proceed  to  the  left  is  determined  by 
the  concentrations  of  C  and  D  and  that  equilibrium  is  reached 
when  the  effects  of  these  two  opposing  changes  exactly  nullify 
each  other. 

A  fairly  exact  quantitative  relationship  exists  among  the  con- 
centrations of  all  the  components  of  a  reversible  reaction  when 
this  reaction  is  at  equilibrium  and  this  is  known  as  the  law  of 
molecular  concentrations,  and  may  be  stated  as  follows:  when  a 
reversible  reaction  has  reached  a  state  of  equilibrium,  the  product 
of  the  molecular  concentrations  of  all  the  components  on  one  side 
of  the  reaction  bears  a  definite  numerical  ratio  to  the  product  of 
the  molecular  concentrations  of  all  the  components  on  the  other 
side  of  the  reaction.  This  ratio  is  known  as  the  equilibrium 
constant  of  the  reaction  and  it  is  always  the  same  at  the  same 
temperature  although  it  may  have  different  values  at  other 
temperatures.  The  word  molecule  here  signifies  any  individual 
component  whether  electrically  charged  or  not.  Thus,  a  solution 
of  acetic  acid  would  contain  the  components  H+,  C2H303~  and 
HC2H3O2.  Concentration  signifies  the  amount  of  the  component 
divided  by  the  volume  and  is  usually  expressed  in  gram  mo- 
lecular weights  per  liter.  For  example,  take  100  cc.  of  a  solution 
containing  6  grams  (0.1  mole)  of  HC2H3O2.  Dividing  the  amount 
(0.1  mole)  by  the  volume,  0.1  liter,  gives  the  ratio  1  and  thus  the 
concentration  of  the  total  acetic  acid  is  1  F.  W.  (formula  weight) 


LAW  OF  MOLECULAR  CONCENTRATION       107 

per  liter.  Since  the  fraction  ionized  is  0.014  the  concentration  of 
hydrogen  ion  [H+]*  is  0.014;  the  concentration  of  the  acetate  ions 
[C2H302~]  is  also  0.014  and  the  concentration  of  un-ionized  acetic 
acid  [HC2H302]  is  1  -  0.014  =  0.986. 

Derivation  of  the  Law  of  Molecular  Concentration  from  a  Con- 
sideration of  the  Speed  of  Reaction.  Let  us  consider  again  the 
general  reaction 

A  +  B  ^±  C  +  D. 

It  is  obvious  that  for  a  molecule  of  A  to  react  with  a  molecule  of 
B  the  two  must  come  into  contact  or  collide.  The  chance  for 
collision  of  any  single  molecule  of  A  with  molecules  of  B  is  pro- 
portional to  the  number  of  B  in  a  given  volume,  that  is,  to  the 
concentration  of  B,  but  there  are  a  great  many  molecules  of  A 
each  of  which  has  the  same  chance  to  collide  with  molecules  of  B. 
Therefore  the  total  number  of  collisions  is  proportional  to  the 
product  of  the  concentrations  of  A  and  B.  But  the  velocity  of 
the  reaction  towards  the  right,  that  is  the  amount  changed  in 
unit  time  in  unit  volume,  is  proportional  to  the  number  of 
collisions.  Thus 

Velocity  (->)  =  fci  [A]  [B]. 

The  factor  ki  is  a  definite  numerical  value  known  as  the  propor- 
tionality constant. 

As  the  reaction  progresses  and  C  and  D  accumulate  it  is  ob- 
vious that  collisions  between  C  and  D  will  ensue,  and  an  exactly 
similar  consideration  will  show  that  the  velocity  in  the  opposite 
direction  is  given  by  the  expression 

Velocity  (<-)  =  A*  [C]  [D] 

in  which  &2  is  likewise  a  constant  which  depends  on  the  chemical 
affinities  of  C  and  D  and  naturally  has  a  different  numerical  value 
than  ki. 

Now  a  reversible  reaction  is  at  a  point  of  equilibrium  when 
no  further  apparent  change  is  taking  place.  The  two  opposing 
reactions  are  without  doubt  taking  place  just  the  same,  but  they 

*  In  mathematical  equations  the  formula  enclosed  in  brackets  signifies 
the  concentration  of  the  substance. 


108  THEORY  OF  IONIZATION 

exactly  undo  the  effect  of  each  other  making  the  total  change 
zero.     Therefore 

vel  (-*)  =  vel  (<-) 
h  [A]  [B]  =  k2  [C]  [D] 

• 


K  is  known  as  the  equilibrium  constant  of  the  reaction.  It  is  the 
ratio  of  the  two  velocity  constants  ki  and  kz.-  - 

The  number  of  components  taking  part  in  a  reversible  reaction 
is  not  always  four  as  in  the  equation  involving  ABCD.  Thus  in 
the  ionization  of  acetic  acid 

HAc^±H+  +  Ac- 

there  is  but  one  component  on  the  left.  Each  molecule  of  HAc 
has  a  definite  tendency  to  ionize  which  is  not  dependent  on  any 
other  dissolved  molecules  or  ion.  Therefore  the  amount  of  acetic 
acid  which  will  ionize  in  unit  time  depends  solely  on  the  amount 
present  or 

vel  (->)  =  fc  [HAc]. 

The  speed  of  the  opposing  reaction  is  given  by 
vel  (<-)  =  k2  [H+]  [Ac-] 

and  the  condition  of  equilibrium  is  given  in  the  expression 

[H+]  X  [Ac-] 
[HAc] 

K  is  the  ionization  constant  of  acetic  acid  and  has  a  numerical 

value  of  0.000018  when  concentrations  are  given  in  moles  per  liter. 

Another  example  of  equilibrium  is  that  between  sulphur  triox- 

ide  and  its  dissociation  products  at  a  temperature  above  500°. 

2SO3  ^  2SO2  +  O2 

The  velocity  towards  the  right  is  dependent  on  the  number  of 
collisions  of  S03  molecules  with  each  other  or 

.  vel  (-»)  =  h  [SO.]  [SO,]  =  A*  [S03]2. 

The  velocity  toward  the  left  is  determined  by  the  number  of  col- 
lisions of  three  different  molecules  or  one  oxygen  molecule  with 
two  S02  molecules 

vel  (<-)  =  k,  [02]  [SOa]  [SQJ 
=  fe  [02]  [S02]2 


LAW  OF  MOLECULAR  CONCENTRATION  109 

and  the  equilibrium  condition  is  determined  by  the  expression 

[S02?  •  [02] 
[S03]2 

Interesting  as  the  foregoing  line  of  reasoning  is,  and  logical  as 
the  deductions  seem  to  be,  the  scientists  who  thought  it  out  would 
at  once  have  discarded  it  if  they  had  not  found  that  it  agreed  to 
a  considerable  degree  of  accuracy  with  the  actual  conditions  found 
to  exist  in  systems  in  equilibrium. 

Thus  actual  measurements  of  the  three  components  in  solu- 
tions containing  acetic  acid,  hydrogen  ions,  and  acetate  ions  show 
that,  whatever  the  actual  concentrations,  the  ratio  always  has  the 
same  value,  namely  that  of  the  equilibrium  constant. 

The  importance  of  the  manufacture  of  sulphuric  acid  has 
caused  a  great  amount  of  study  to  be  made  of  the  equilibrium 
between  sulphur  dioxide,  oxygen  and  sulphur  trioxide,  and  the 
validity  of  the  law  of  molecular  concentrations,  as  applied  to 
this  reaction,  has  been  put  to  a  rigid  test. 

The  law  is  in  fact  of  very  wide  application;  it  holds  for  non- 
ionic  as  well  as  ionic  reactions.  The  degree  of  ionization  of 
weakly  ionized  substances  can  be  figured  with  high  precision  ac- 
cording to  the  law.  But  the  behavior  of  strong  electrolytes 
does  not  conform  as  closely  to  this  law,  and  the  law  is  of  value 
only  in  a  qualitative  fashion  to  predict  the  extent  of  the  ionization 
of  these  substances.  In  this  connection  we  may  recall  Rule  4 
for  writing  ionized  equations,  which  directed  to  treat  all  strong 
electrolytes  as  if  they  were  completely  ionized.  In  many  re- 
spects all  strong  electrolytes  do  actually  behave  in  solution  as  if 
they  were  in  fact  completely  ionized.  Some  chemists  are  inclined 
to  believe  that  this  is  really  true,  and  that  the  methods  used  for 
measuring  degree  of  ionization,  —  electrical  conductivity  and 
freezing  point,  —  by  which  the  figures  printed  in  the  table  in  this 
chapter  are  determined,  do  not  really  give  the  true  results.  Be 
this  as  it  may,  there  is  good  justification  for  employing  the  ap- 
proximation of  considering  strong  electrolytes  as  completely 
ionized,  when  no  great  degree  of  accuracy  is  needed.  On  the  other 
hand  the  degree  of  ionization  of  weak  electrolytes  follows  with 
a  high  degree  of  accuracy  the  law  of  molecular  concentra- 
tion. 


110  THEORY  OF  IONIZATION 

PROBLEMS 

87.  Acetic  acid  in  0.1  molal  solution  is  1.4  percent  ionized. 
Find  the  value  of  the  ionization  constant  K. 

HAc^±H+  +  Ac- 

[H+]:5<  [Ac-] 
[HAc] 

88.  Find  the  percent  ionization  of  acetic  acid  in  molal 
solution.     Take   the   value   of    the    ionization    constant   as 
0.000018  and  solve  the  quadratic  equation  by  the  method 
of  approximation. 

89.  What  is  the  concentration  of  hydrogen  ions  in  a  0.1 
molal  solution  of  acetic  acid? 

90.  What  is  the  concentration  of  hydrogen  ions  in  a  solu- 
tion containing  in  a  liter  0.1  mole  of  acetic  acid  and  0.1  mole 
of  NaAc?    Assume  as  an  approximation  that  the  salt  is  100 
percent  ionized. 

91.  For  the  sake  of  showing  the  different  effects  of  a  neu- 
tral salt  of  a  strong  acid  upon  the  ionization  of  the  acid,  find 
first  the  ionization  constant  of  nitric  acid  if  it  is  90  percent 
ionized  in  0.1  molal  solution.     Then  using  the  constant  thus 
found  calculate  the  hydrogen  ion  concentration  in  a  solution 
containing,  in  1  liter,  0.1  mole  of  HNO3  and  0.1  mole  of  KN03. 
Assume  in  this  calculation  that  the  law  of  molecular  con- 
centration   holds    accurately.     The    result    shows    that   the 
effect  of  a  neutral  salt  of  a  strong  acid  upon  the  ionization 
of  the  acid  is  not  marked  as  it  is  in  the  similar  case  of  a  weak 
acid.     As  a  matter  of  fact  the  actual  effect  is  even  less  than 
that  calculated  according  to  the  law. 

92.  The  ionization  constant  of  ammonium  hydroxide  is 
0.000018.     (a)  Find   the   concentration   of   OH~  ions  in  a 
molal  solution  of  NH4OH.     (6)  Find  the  concentration  of 
OH~  ions  in  a  molal  solution  of  NH^OH  which  contains 
also  0.5  moles  of  NH4C1  per  liter  assuming  the  latter  to  be 
completely  ionized. 

93.  What  is  the  ionization  constant  of  hydrocyanic  acid 
if  the  ionization  in  0.1  normal  solution  is  0.01  percent? 

94.  What  is  the  hydrogen  ion  concentration  in  a  solution 
containing,  in  1  liter,  0.1  mole  of  HCN  and  0.1  mole  of  KCN 
assuming  the  latter  to  be  100  percent  ionized? 


SOLUBILITY  AND  SOLUBILITY  PRODUCT  111 

SOLUBILITY  AND  SOLUBILITY  PRODUCT 

The  dissolving  of  a  substance  has  been  variously  considered  as 
a  physical  and  a  chemical  change,  but  at  all  events  it  is  a  re- 
versible change  which  like  reversible  chemical  reactions  is  sub- 
ject to  the  law  of  molecular  concentrations. 

When  a  solid  non-electrolyte  such  as  urea  is  placed  in  water 
it  begins  to  dissolve  according  to  the  reaction 

CO(NH2)2  ->  CO(NH2)2 

but,  as  the  urea  accumulates  in  solution,  the  reverse  reaction  be- 
comes more  and  more  rapid  until  its  speed  is  just  equal  to  the  rate 
of  solution,  at  which  point  equilibrium  is  established.  Applying 
the  expression  for  equilibrium 

[CO(NH2)2] 


[CO(NH2)2] 


=  constant 


we  arrive  at  a  conclusion  which  is  quite  in  accord  with  the  facts. 
For  the  concentration  of  the  solid  urea  is  a  perfectly  definite  and 
unchangeable  quantity,  in  other  words  it  is  a  constant;  then 
by  transposing  the  equation  we  obtain 

[CO(NH2)2]  =  [CO(NH4)2]  X  constant 
=  constant2 

This  is  simply  an  expression  of  the  well-known  fact  that  the  solu- 
bility of  urea  is  a  perfectly  definite  quantity  for  a  definite  tem- 
perature. It  should  be  particularly  noticed  that  the  solubility 
depends  in  no  way  on  the  quantity  of  solid  material  left  in  con- 
tact with  the  solution.  Equilibrium  can  exist  as  well  between  a 
single  small  fragment  of  the  solid  and  the  solution  as  it  can  when 
there  is  a  large  amount  of  the  solid  present. 

Now  that  we  have  considered  the  simplest  case  of  equilibrium 
between  a  solid  and  its  solution  we  are  in  a  position  to  understand 
the  more  complex  case  in  which  the  dissolved  substance  is  ionized. 
Let  us  consider  for  example  the  sparingly  soluble  silver  bromate 

AgBrO3  ^  Ag+  +  BrO3- 

If  we  consider  the  soluble  portion  to  be  entirely  ionized  as  a 
justifiable  approximation  when  only  moderate  accuracy  is  desired, 
it  is  apparent  that,  to  add  a  particle  to  the  crystal,  an  Ag+  ion  and 
a  Br03~  ion  must  collide  at  the  same  instant  at  the  same  spot 


112  THEORY  OF  IONIZATION 

of  the  crystal  surface.  The  velocity  of  crystallization  is  there- 
fore proportional  to  the  product  of  the  concentration  of  the  two 
ions 

Vel  (<-)  =  fa  X  [Ag+]  X  [BrO,-] 

whereas  the  velocity  of  solution  is  proportional  to  the  fixed  quan- 
tity, the  concentration  of  the  solid  AgBrO3  of  the  crystal 

Vel  (->)  =  fa  X  [AgBrO,]. 
The  equilibrium  constant  therefore  is  given  by  the  expression 

[Ag+]  X  [BrO,-]  _ 
[AgBrOs] 

or  since  [AgBrQ3]  is  constant 

[Ag+]  X  [Br03~]  =  K2  =  solubility  product. 

This  is  known  as  the  solubility  product. 

Let  us  consider  a  solid  salt  which  gives  three  ions  in  solution, 
lead  iodide  for  example 


In  this  case  three  ions  must  collide  simultaneously  at  the  surface 
of  the  crystal  to  produce  a  building  up  and  the  velocity  of  crystal- 
lization is 

Vel  (<-)  =  fa  X  [Pb++]  X  [I-]  X  [I-] 
=  fa  X  [Pb++]  X  [I-]2 

Whereas  the  velocity  of  solution  is  again  proportional  to  the  fixed 
concentration  of  the  solid  crystal  or 

Vel  (->)  =  fa  X  [Pblj. 

Therefore  the  solubility  product  is  given  by  the  expression 
[Pb++]  X  [I-]2  =  K2  =  solubility  product. 

It  has  already  been  stated  that  the  ionization  of  strong  electro- 
lytes in  solution  does  not  conform  closely  to  the  law  of  molecu- 
lar concentrations.  However  in  the  case  of  saturated  solutions 
the  equilibrium  between  the  solid  salt  and  its  ions  in  solution  does 
conform  very  closely  to  the  general  law  and  this  specific  case  is 
known  as  the  principle  of  solubility  product. 

In  a  saturated  solution  of  any  electrolyte  the  product  of  the  con- 
centrations of  its  ions  is  a  constant  at  any  given  temperature.     This 


SOLUBILITY  AND  SOLUBILITY  PRODUCT  113 

principle  is  of  great  value  in  calculating  the  solubility  of  any  given 
salt  in  a  solution  containing  another  salt  with  an  ion  in  common 
with  the  given  salt. 

An  example  will  make  the  application  of  this  principle  clear. 

Let  us  prepare  a  saturated  solution  of  silver  acetate  by  shaking 
some  of  the  solid  salt  for  a  long  time  with  pure  water.  Let  the 
surplus  salt  settle,  then  pour  a  little  of  the  clear  solution  into  a 
small  beaker.  Now  drop  a  clear  lump  of  sodium  acetate  into  the 
beaker.  The  latter  salt  is  excessively  soluble  and  will  have  com- 
pletely dissolved  within  a  few  moments.  Very  shortly  there  is 
seen  a  cloud  of  fine  flaky  crystals  settling  from  all  parts  of  the 
solution.  This  clearly  is  a  different  substance  from  the  sodium 
acetate  added,  and  tests  upon  it  easily  show  that  it  is  silver  acetate. 

The  solution  was  already  saturated  with  silver  acetate.  The 
excess  of  acetate  ions  supersaturated  the  solution  and  the  solid 
separated  according  to  the  reaction 

Ag+  +  Ac~  — >  AgAc  I 

until  the  solution  was  left  exactly  saturated,  at  which  point  the 
product  of  the  concentrations  of  the  two  ions  was  exactly  equal 
to  the  solubility  product.  Naturally,  under  this  new  condition 
of  equilibrium,  there  are  many  more  acetate  ions  than  silver  ions 
in  solution,  but  the  product  is  equal  to  the  same  value  as  in  a 
solution  of  silver  acetate  alone,  in  which  the  concentration  of  the 
two  ions  is  the  same. 

If  a  little  more  of  the  saturated  solution  of  silver  acetate  is 
poured  into  another  beaker  and  a  small  crystal  of  silver  nitrate  is 
dropped  in,  the  same  thing  occurs.  The  silver  nitrate  is  ex- 
cessively soluble  and  entirely  dissolves  in  a  few  moments.  Very 
soon  a  similar  flaky  precipitate  of  silver  acetate  crystals  falls  but. 

PROBLEMS 

95.  The  molal  solubility  of  AgBrO3  at  24.5°  is  0.0081. 
How  many  grams  of  AgBr03  could  dissolve  in  a  liter  of  0.1 
molal  AgNO3?    Assume  complete  ionization  of  the  salts. 

96.  The  solubility  of  PbSO4  is  0.04  gram  per  liter  at  25°. 
How  many  grams  of  sodium  sulphate  should  be  added  to 
1  liter  to  reduce  the  solubility  of  the  lead  sulphate  to  0.0001 
grams  per  liter? 


CHAPTER  IV 

THE  NON-METALLIC  ELEMENTS  IN  BINARY 
COMPOUNDS 

The  distinctive  chemical  property  of  non-metallic  elements  is 
their  ability  to  combine  with  metals  forming  simple  compounds 
in  which  they  are  the  negative  constituent.  It  is  the  purpose  of 
this  chapter  to  deal  with  such  simple  compounds  of  the  more 
pronounced  non-metals:  fluorine,  chlorine,  bromine,  iodine,  oxy- 
gen, sulphur,  and  nitrogen. 

PREPARATION  4 
COPPER  OXIDE 

This  preparation  is  typical  of  the  preparation  of  oxides  of  the 
metals.  The  most  obvious  method  of  making  an  oxide  is  to  heat 
the  substance  in  air  or  oxygen,  but  the  difficulty  with  such  a 
method  is  that  the  oxide  is  solid  and  adheres  to  the  surface,  pre- 
venting the  action  on  the  metal  underneath. 

In  the  method  here  used  the  metal  is  treated  with  a  reagent 
which  yields  a  soluble  compound.  The  latter  is  obtained  by 
evaporating,  and  is  then  decomposed,  by  heating,  into  a  gaseous 
oxide  and  the  solid  oxide  of  the  metal. 

Materials:      Copper  metal    (turnings  or  wire),    13  grams  = 

£  gram  atom. 

6-normal  HN03,  89  cc. 

anhydrous  sodium  carbonate,  21  grams. 
Apparatus:    2  liter  common  bottle. 

5-inch  filter  funnel  with  plain  filter. 

4-inch  evaporating  dish. 

Procedure:  Dissolve  the  copper  in  the  nitric  acid  observing 
Note  7,  page  13.  Add  500  cc.  water  and  filter  if  the  solution 
is  not  clear.  Dissolve  the  sodium  carbonate  in  500  cc.  water, 
filter  if  the  solution  is  not  absolutely  clean  and  pour  this  solution 
into  the  copper  nitrate  solution  in  a  2  liter  common  bottle.  The 

114 


BARIUM  PEROXIDE  HYDRATE  115 

precipitate  of  basic  copper  carbonate 'is  now  to  be  washed  free 
of  the  soluble  sodium  nitrate.  Wash  by  decantation  (Note  5(6), 
page  10)  until  not  more  than  one  one-thousandth  of  the  original 
NaNOs  is  left.  Then  pour  the  sludge  of  basic  copper  carbonate 
in  a  filter  folded  in  a  5-inch  funnel  and  let  it  drain  and  partially 
dry  until  it  has  stiffened  to  a  jelly.  Then  remove  the  filter  and 
contents,  and,  without  tearing  the  filter,  spread  it  out  flat  on  some 
paper  towels.  When  the  copper  carbonate  is  dry  place  it  in  a 
4-inch  evaporating  dish  and  warm  it  by  playing  the  flame  under- 
neath. A  very  low  temperature  suffices  to  decompose  the  copper 
carbonate.  When  the  substance  has  changed  to  a  uniform  black 
powder  and  no  more  gas  is  evolved  the  preparation  is  complete. 
Put  it  up  in  a  6-inch  cork  stoppered  test  tube. 

QUESTIONS 

1.  Could  hydrochloric  acid  take  the  place  of  nitric  acid  in  this 
preparation?     Explain. 

2.  Refer  back  to  Experiment  1,  Chapter  I  for  the  method  of 
obtaining  zinc  oxide  from  zinc.     What  would  be  the  advantages 
and  disadvantages  of  using  that  method  in  this  preparation? 

PREPARATION  5 

r       HYDROGEN  PEROXIDE  AND  BARIUM  PEROXIDE  HYDRATE 

The  ordinary  oxide  of  barium  has  the  formula  BaO  and  it  is  an 
ordinary  metal  oxide,  that  is,  it  forms  the  corresponding  base 
Ba(OH)2  when  it  combines  with  water,  and  the  ordinary  barium 
salts  are  formed  when  this  base  is  neutralized  with  acids.  But 
there  is  another  oxide  of  barium  containing  twice  as  much  oxy- 
gen, BaO2  which  is  easily  obtained  when  barium  oxide  is  heated 
to  a  dull  red  heat  in  air  free  from  carbon  dioxide. 

This  oxide  however  is  not  an  ordinary  oxide;  it  does  not  for 
example  react  with  water  to  give  such  a  base  as  Ba(OH)4  nor  with 
acids  to  yield  such  a  salt  as  Ba(NO3)4.  The  entire  oxygen  seems 
to  behave  as  a  single  radical  having  the  same  valence,  namely  two, 
as  the  simple  oxygen  radical  in  ordinary  barium  oxide;  because, 
when  barium  peroxide  is  treated  with  acids,  the  ordinary  barium 
salt  is  formed  and  hydrogen  peroxide,  e.g. 

BaO2  +  2HC1  -+  BaCl2  +  H202 


116     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

The  name  peroxide  has  been  given  to  such  oxides  containing  the 
O2  radical  which  is  interchangeable  with  the  negative  radicals  of 
acids  so  as  to  yield  hydrogen  peroxide. 

The  common  heavy  metals  have  not  the  ability  to  form  perox- 
ides; only  the  most  strongly  metallic  elements  such  as  barium 
and  the  alkali  metals  do  so.  For  example,  when  sodium  metal 
burns  freely  in  air  the  peroxide  Na^  is  formed.  It  is  a  matter 
of  a  good  deal  of  difficulty  to  limit  the  access  of  oxygen  to  burning 
sodium  so  as  to  obtain  the  ordinary  oxide  Na2O. 

In  the  following  preparation  crude  barium  peroxide  is  treated 
with  an  acid  to  obtain  a  solution  of  hydrogen  peroxide.  After 
removal  of  impurities  by  filtration,  a  solution  of  barium  hydroxide 
is  added  whereby  pure  crystalline  barium  peroxide  hydrate 
BaO2.8H20  is  precipitated. 

Hydrogen  peroxide  may  be  regarded  as  a  weak  acid.  It  is 
excessively  weak,  being.no  more  ionized  than  water  itself.  Yet  its 
acid  character  is  manifest  from  the  way  it  reacts  with  the  barium 
hydroxide : 

Ba(OH)2  +  H2O2  ->  Ba02 j  +  2H20. 

Were  the  barium  peroxide  soluble  this  reaction  would  shortly 
come  to  equilibrium,  but  since  the  barium  peroxide  is  very  in- 
soluble, crystallizing  as  the  hydrate,  the  reaction  is  fairly  com- 
plete. 

Materials:       crude  BaO2     34  grams  =  ^  F.W. 

6-normal  HC1     67  cc.  =  |  F.W. 

barium  hydroxide  Ba(OH)2.8H2O    63  grams  =  £ 
F.W. 

ice. 
Apparatus:     2-600  cc.  flasks. 

5-inch  filter  funnel  with  plain  filters. 

2-liter  common  bottle. 

600  cc.  beaker. 

suction  filter. 

Procedure:  Dissolve  the  barium  hydroxide  in  500  cc.  of  warm 
water  and  pour  through  a  filter  without  suction  into  a  600  cc. 
flask.  While  the  filter  is  draining  place  the  HC1,  250  cc.  ice 
water,  and  about  100  grams  ice  in  a  600  cc.  beaker.  Grind  the 
BaO2  in  a  mortar  with  water  until  a  smooth  uniform  paste  is  ob- 


HYDROCHLORIC  ACID  117 

tained  and  add  enough  water  to  make  100  cc.  Add  the  Ba(>2 
suspension  a  little  at  a  time  with  constant  stirring  to  the  cold 
HC1  solution  until  the  Ba02  ceases  to  dissolve,  or  until  it  is  all 
added.  Now  add  a  few  cc.  of  the  Ba(OH2)  until  a  light  pre- 
cipitate is  formed.  This  will  be  colored  reddish  due  to  ferric 
hydroxide.  The  solution  should  be  slightly  alkaline  at  this  point. 
Make  sure  of  this  by  testing  with  litmus  paper.  Then  filter  the 
solution  without  suction  into  a  600  cc.  flask.  Pour  the  Ba(OH)2 
solution  into  a  2-liter  bottle,  and  add  1  liter  of  cold  water.  Then 
pour  in  a  thin  stream  with  constant  stirring  the  hydrogen  peroxide 
solution  into  the  barium  hydroxide  solution.  Let  the  flaky 
barium  peroxide  hydrate  settle  and  then  collect  it  on  a  suction 
filter.  As  soon  as  the  water  is  drawn  out,  shut  off  the  suction, 
wash  with  15  cc.  cold  water,  and  again  suck  dry.  Do  not  draw 
any  quantity  of  air  through  the  product.  Wrap  the  crystals  in 
paper  towels  and  dry  them  according  to  Note  9(6),  page  15. 
Preserve  the  product  in  a  4-ounce  cork  stoppered  bottle. 

QUESTIONS 

1.  Explain  why  barium  chloride  would  not  give  a  precipitate 
with  hydrogen  peroxide  as  well  as  barium  hydroxide. 

2.  The  hydrogen  peroxide  solution  obtained  as  an  intermediate 
product  in  this  preparation  contained  barium  chloride.     Suggest 
with  what  reagent  one  might  treat  the  purified  barium  peroxide 
hydrate  to  obtain  a  pure  solution  of  hydrogen  peroxide.     Give 
explanation. 

PREPARATION  6 
HYDROCHLORIC  ACID 

Sulphuric  acid  and  sodium  chloride  are  the  sources  respectively 
of  the  positive  and  negative  radical  in  this  preparation.  The 
hydrogen  and  the  chlorine  radical  combining  give  hydrogen  chlo- 
ride, the  only  volatile  product  possible,  and  this  is  led  off  to  be 
dissolved  in  water  in  another  part  of  the  apparatus. 

Materials:       rock  salt,  NaCl,  117  grams  =  2  F.W. 

36-normal  sulphuric  acid,  222  cc. 
Apparatus:     1500  cc.  round  bottom  flask. 

1   wide   mouth   bottle   of   300-500  cc.   capacity 


118      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

with  as  wide  a  mouth  as  possible  (or  better  a 

3' necked  Woulfe  bottle). 
1  common  300  cc.  bottle. 
1  thistle  tube. 

1  2-hole  rubber  stopper  to  fit  flask. 
1  3-hole  rubber  stopper  to  fit  first  bottle, 
glass  delivery  tube  fitted  as  in  diagram, 
drying  tube  packed  with  glass  wool. 
1  1-hole  rubber  stopper  to  fit  drying  tube, 
pan  of  cold  water  in  which  first  absorption  bottle 

is  to  set. 

Procedure:  Study  the  diagram  of  the  apparatus  thinking  out 
the  purpose  of  each  part  so  that  it  may  be  put  together  intelli- 
gently. Build  the  apparatus.  Place  the  dry  rock  salt  in  the 


Wool 


Water 


EIG.  15 


flask,  and  167  cc.  of  water  in  each  absorption  bottle.  Adjust  the 
inlet  tube  of  first  absorption  bottle  so  that  it  reaches  to  within 
J  inch  of  the  surface  of  the  water. 

Pour  enough  of  the  sulphuric  acid,  perhaps  50  cc.,  through  the 
thistle  tube  into  the  flask  to  cover  the  lower  end  of  the  thistle 


HYDROCHLORIC  ACID 


119 


tube.  Watch  the  action  carefully  to  see  that  the  foam  does  not 
rise  and  threaten  to  go  up  into  the  neck  of  the  flask.  As  rapidly 
as  seems  safe  add  the  rest  of  the  sulphuric  acid.  When  the  action 
slackens  apply  heat  very  cautiously  to  the  flask  and  finally  heat 
it  strongly  until  all  the  salt  has  dissolved  and 
effervescence  has  ceased. 

Danger:  The  contents  of  the  flask  are  very 
hot  and  must  not  be  poured  out,  least  of  all 
into  water.  Let  the  contents  cool  completely 
(best  until  next  exercise),  then  add  cold  water 
and  wash  out  the  flask. 

Most  of  the  hydrogen  chloride  has  been  dis- 
solved in  the  first  bottle,  unless  the  latter  was 
allowed  to  get  too  hot,  and  any  remainder  has 
been  caught  in  the  second  bottle. 

To  find  the  concentration  of  the  acid  in  the 
first  bottle  determine  the  specific  gravity  using 
a  Westphal  Balance,  or  a  hydrometer,  or  the 
apparatus  shown  in  the  diagram. 

Look  up  in  a  specific  gravity  table  the 
normality  corresponding  to  the  specific  gravity, 
measure  the  total  volume  of  the  acid  and 
compute  the  number  of  formula  weights  and 
the  number  of  grams  of  HC1  obtained  in  the  first 
bottle.  If  a  satisfactory  yield  cannot  be  ac- 
counted for,  determine  in  the  same  way  the 
acid  in  the  second  bottle.  The  contents  of 
this  however  is  not  to  be  saved.  Preserve 


100 


FIG.  16 


FIGURE  16. 

The  scale  at  the  back  is  marked  off  in  divisions  of  ^  inch.  The  acid  is 
placed  in  the  shallow  dish  at  the  left  and  water  in  the  dish  at  the  right, 
both  being  at  the  same  temperature.  The  level  is  adjusted  in  both 
dishes  so  that  the  surface  of  each  liquid  is  exactly  at  the  zero  point  of 
the  scale.  Suction  is  applied  at  the  top  until  the  acid  rises  exactly  to 
the  100  mark  when  the  pinch  cock  is  closed.  The  reading  of  the  scale 
at  the  top  of  the  water  column  divided  by  100  gives  the  specific  gravity 
of  the  acid. 


the  acid  from  the  first  bottle  in  a  250  cc.  narrow-necked  glass- 
stoppered  bottle. 


120      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

QUESTIONS 

1.  How  much  of  an  excess  of  sulphuric  acid  was  used  in  this 
preparation?  What  is  the  advantage  of  using  such  an  excess? 

PREPARATION  7 
HYDROBROMIC  ACID 

The  method  of  the  preceding  preparation  would  be  the  obvious 
one  for  making  hydrobromic  acid  were  it  not  for  the  fact  that 
hydrogen  bromide  reacts  to  some  extent  with  concentrated  sul- 
phuric acid  according  to  the  equation 

H2SO4  -+  2HBr  ->  2H2O  +  SO2  +  Br2 

and  the  gas  passing  over  to  the  absorption  bottles  would  be  con- 
taminated with  sulphur  dioxide  and  bromine. 

Hence  the  method  we  shall  choose  starts  with  bromine  and 
allows  this  to  react  with  red  phosphorus  and  water.  Phosphorus 
and  bromine  combine  very  easily  to  form  PBr3;  but  the  phos- 
phorus bromide  hydrolyzes  at  once  completely: 

PBr3  +  3H2O  -4  P(OH)3  +  3HBr 
and  hydrogen  bromide  passes  over  to  the  absorption  flask. 

Materials: .     bromine,  40  grams  =  12.5  cc.  =  J  F.W. 

red  phosphorus,  7  grams. 
Apparatus:     dropping  funnel. 

250  cc.  flask. 

4-inch  side  arm  U-tube. 

8-inch  plain  U-tube  of  f  to  1-inch  bore. 

1  1-hole  rubber  stopper  to  fit  flask. 

2  solid  rubber  stoppers  to  fit  small  U-tube. 
1  1-hole  rubber  stopper  to  fit  large  U-tube. 
glass  delivery  tube  as  in  diagram. 

5  grams  broken  glass, 
glass  wool. 

Procedure:  After  carefully  studying  the  diagram  construct 
the  apparatus.  Place  6  grams  of  the  phosphorus,  12  grams  of 
sand  and  6  cc.  of  water  in  the  flask.  Mix  1  gram  of  the  phos- 
phorus with  the  broken  glass,  place  it  in  the  bottom  of  the 
small  U-tube,  and  press  into  each  arm  above  it  a  small  wad  of  glass 


HYDROBROMIC  ACID 


121 


wool.     Add  12  cc.  water,  or  enough  to  close  the  bend,  to  the  larger 
U-tube. 

Danger:  Use  extreme  care  in  handling  bromine  as  it  produces 
very  severe  burns  in  the  flesh.  Be  sure  the  stop  cock  of  the  drop- 
ping funnel  is  tight  and  turns  easily.  When  manipulating  it  after 


•  Ring  Stand - 

FIG.  17 

the  bromine  is  added,  steady  the  bulb  with  one  hand  while  turn- 
ing the  cock  with  the  other  and  be  very  careful  that  the  cock 
does  not  slip  out  of  the  socket  letting  the  bromine  leak  out  over 
the  fingers. 

After  making  sure  that  the  apparatus  is  tight  pour  the  bromine 
into  the  funnel.  Then  let  a  single  drop  of  bromine  fall  into  the 
flask  watching  its  effect  then  add  the  rest  of  the  bromine  a  drop 
at  a  time  as  rapidly  as  proves  safe.  After  the  gas  ceases  to  flow 
pour  the  contents  of  the  absorption  tube  into  a  graduate,  note  its 
volume,  and  preserve  it  in  a  25  cc.  glass-stoppered  bottle.  Wash 
out  the  apparatus  at  the  sink  under  the  hood. 


QUESTIONS 


1.  Why  would  not  a  metal  like  zinc,  which  will  combine  vig- 
orously with  bromine  serve  instead  of  phosphorus? 

2.  Explain  why  phosphorus  is  placed  in  the  small  U-tube. 


122      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

3.  Explain  why  at  the  outset  of  the  process  bubbles  pass  out- 
ward through  the  second  U-tube,   and  later  they  pass  in  the 
opposite  direction. 

4.  What   are   the   products  of   the   reaction   of   concentrated 
sulphuric  acid  with  potassium  bromide?     (See  Exp.  13.)     Write 
the  equations.     What  property  possessed  by  hydrogen  bromide, 
but  not  by  hydrogen  chloride,  is  shown? 

PREPARATION  8 
HYDRIODIC  ACID 

The  method  employed  in  the  preparation  of  hydrochloric  acid 
is  inapplicable  here  for  the  same  reason  that  it  is  in  the  preparation 
of  hydrobromic  acid.  v  Hydriodic  acid  can  be  prepared  by  the 
same  method  which  we  used  for  hydrobromic  acid,  that  is,  by 
letting  iodine,  phosphorus,  and  water  react;  but  in  order  to  illus- 
trate a  different  principle  we  shall  employ  another  method. 

The  position  of  sulphur  in  the  negative  electromotive  series 
is  lower  than  that  of  iodine  because  the  latter  displaces  sulphur 
from  a  solution  of  hydrogen  sulphide.  This  is  the  principle  on 
which  this  preparation  is  based. 

Materials:  iodine,  32  grams  =  f  gram  atomic  weight.  (Keep 
the  iodine  stoppered  in  a  test  tube  until  ready 
to  use.) 

hydrogen  sulphide  generator,  see  Note  13,  page  19. 
Apparatus:    200  cc.  Erlenmeyer  flask, 
two  75  cc.  flasks. 
150  cc.  distilling  flask, 
thermometer, 
condenser. 

carbon  dioxide  generator, 
hydrogen  generator. 

cork  or  rubber  stoppers  to  fit  condenser  to  dis- 
tilling flask  and  receiving  flask. 

Procedure:  Grind  the  iodine  to  a  fine  powder  in  a  mortar,  and 
quickly  replace  it  in  the  stoppered  test  tube.  Place  about  0.5 
gram  iodine  and  100  cc.  water  in  the  Erlenmeyer  flask.  Pass 
hydrogen  sulphide  into  the  solution  until  the  brown  color  of  iodine 
has  disappeared.  Add  about  1  gram  more  of  iodine  and  again 


HYDRIODIC  ACID  123 

pass  hydrogen  sulphide  until  the  iodine  is  used  up.  After  10 
grams  of  iodine  have  reacted  in  this  way,  add  the  remaining  20 
grams  and  allow  the  mixture  to  stand,  with  repeated  shaking,  until 
the  iodine  is  entirely  dissolved  (half  an  hour  or  more).  Then 
pass  hydrogen  sulphide  slowly  until  the  solution  is  decolorized. 
Pour  the  solution  into  another  flask,  leaving  the  clotted  lunlps 
of  sulphur  behind,  and  rinse  the  first  flask  and  the  residue  with  a 
few  cc.  of  water.  Pass  a  current  of  carbon  dioxide  through  the 
solution  until  the  excess  of  hydrogen  sulphide  is  entirely  removed; 
then  shake  the  flask  vigorously  to  cause  the  suspended  sulphur  to 
clot  together,  and  filter  the  solution.  In  this  way  a  rather  weak 
solution  of  hydriodic  acid  is  obtained. 

Fit  a  distilling  flask  with  a  thermometer  and  an  inlet  tube  for 
hydrogen,  and  pass  the  side  arm  of  the  .flask  into  a  condenser. 
After  introducing  the  hydriodic  acid  solution,  fill  the  whole  ap- 
paratus with  hydrogen,  and  keep  a  slow  current  of  this  gas  passing 
during  the  distillation.  On  distilling,  nearly  pure  water  passes 
over  at  first,  and  the  thermometer  does  not  register  appreciably 
above  100°.  When  the  thermometer  rises  to  105°  change  the  re- 
ceiving vessel  and  collect  the  distillate  until  the  temperature  has 
risen  to  120°.  Change  the  receiver  again  and  collect  the  rest  of 
the  distillate.  The  temperature  rises  quickly  to  126°,  and  re- 
mains very  close  to  this  point  until  practically  all  of  the  acid  has 
passed  over.  This  last  fraction  is  the  desired  concentrated  acid. 
Preserve  this  acid  in  a  50  cc.  glass-stoppered  bottle. 

QUESTIONS 

1.  Compare  the  stability  of  the  hydrogen  halides  as  judged 
(a)  by  the  heat  produced  when  they  are  formed  from  the  ele- 
ments;  and  (6)  by  the  temperature  at  which  the  compounds  are 
decomposed.     Look  up  the  data. 

2.  What  are  the  products  of  the  reaction  of  concentrated  sul- 
phuric acid  and  potassium  iodide  (Experiment  14)?    Write  the 
equations.     What  property  of  hydrogen  iodide  is  shown?     Com- 
pare hydrogen  iodide,  hydrogen  bromide  and  hydrogen  chloride 
as  reducing  agents. 

3.  Is  sulphur  invariably  a  less  active  element  than  iodine? 
Look  up  the  decomposition  temperatures  of  hydrogen  sulphide 
and  hydrogen  iodide.     Find  in  reference  book  the  effect  of  heating 
sulphur  in  dry  hydrogen  iodide. 


124     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 
4.   What  explanation  can  you  offer  for  the  fact  that  the  reaction 

H2S  +  I2  ^±  2HI  +  S 

goes  in  one  direction  in  aqueous  solutions  and  in  the  other  direc- 
tion in  the  dry  gases'?  Look  pp  the  heats  of  solution  of  HI  and 
H2S. 

PREPARATION  9 
BARIUM  BROMIDE,  BaBr2.2H2O 

A  universal  method  of  preparing  a  salt  is  by  neutralizing  the 
respective  acid  and  base.  It  is  almost  as  effective  to  use  the  car- 
bonate of  the  metal  as  the  base  itself,  because  carbonic  acid  is  so 
easily  displaced.  When  both  the  acid  and  the  base  are  soluble, 
it  is  necessary  to  get  the  solution  exactly  neutral  if  a  perfectly 
pure  salt  is  to  be  made;  but  in  this  preparation  the  barium  car- 
bonate used  to  react  with  the  acid  is  so  very  insoluble  that  it 
may  be  used  in  excess  and  the  excess  may  be  filtered  off.  Thus 
the  tedious  process  of  balancing  the  base  against  the  acid  to  get 
the  exact  neutral  point  is  avoided. 

Materials:      hydrobromic  acid.     Use  product  obtained  in  No.  7. 

barium  carbonate  BaCOs,  50  grams. 
Apparatus:     5-inch  filter  funnel. 

500  cc.  flask. 

4-inch  evaporating  dish. 

Procedure:  Place  all  but  2  grams  of  the  barium  carbonate  in 
the  flask,  add  100  cc.  of  water  and  then  the  hydrobromic  acid 
cautiously  enough  not  to  cause  too  violent  frothing.  When 
even  with  shaking,  effervescence  ceases,  add  the  other  2  grams 
of  barium  carbonate.  If  this  also  goes  into  solution,  add  more 
barium  carbonate  until  an  excess  remains  undissolved.  Boil  the 
contents  of  the  flask  5  minutes.  Filter  without  suction  and 
partially  evaporate  the  filtrate.  Let  the  solution  cool  and 
crystallize.  Collect  the  crystals,  and  evaporate  the  mother 
liquor.  In  this  way  practically  the  entire  amount  of  salt  should 
be  obtained  in  the  crystalline  form.  Wrap  the  product  in  paper 
towels  and  leave  it  to  dry  (Note  9(6),  page  15).  Put  up  the 
product  in  a  2-ounce  cork-stoppered  bottle. 


ALUMINUM  SULPHIDE  125 

QUESTIONS 

1.  Explain  the  method  employed  in  this  preparation  of  obtain- 
ing an  exactly  neutral  salt  solution.     Write  ionized  equation. 

2.  Could  a  neutral  solution  of  barium  bromide  be  obtained  by 
adding  excess  of  barium  hydroxide  to  hydrobromic  acid?     Ex- 
plain. 

3.  Could  a  neutral  potassium  salt  solution  be  obtained  by 
treating  the  appropriate  acid  with  excess  of  potassium  carbonate? 

PKEPARATION  10 
ALUMINUM  SULPHIDE 

Although  the  obvious  method  to  prepare  a  binary  compound 
would  seem  to  be  to  bring  the  two  elements  together,  such  a 
procedure  is  not  very  often  followed.  Aluminum  and  sulphur 
can  be  made  to  combine  directly,  it  is  true,  but  the  process  is 
so  difficult  to  control  that  we  shall  not  attempt  it.  When  finely 
divided  aluminum  and  sulphur  are  mixed  and  the  mixture  is 
heated,  either  the  sulphur  entirely  distils  off  without  any  re- 
action taking  place,  or  if  a  reaction  starts  it  is  too  violent  to  con- 
trol. 

We  have  therefore  selected  lead  sulphide  as  a  source  of  the  sul- 
phide radical  because  this  substance  is  not  volatile  and  cannot 
escape  before  it  reacts;  and,  when  it  does  react,  the  reaction  is  not 
too  violent,  because  a  part  of  the  energy  furnished  by  the  com- 
bination of  aluminum  and  sulphur  is  expended  in  separating  lead 
and  sulphur. 

Materials:      granulated  aluminum,  Al,  27  grams  =  1  F.W. 
|  ,..  pulverized  galena,  lead  sulphide,  PbS,  359  grams 

=  1.5  F.W.  (a  fairly  pure  sample  of  the  min- 
eral must  be  used  to  obtain  good  results). 
Apparatus:    clay  crucible  without  cover, 
iron  pan. 
gas  furnace. 

Procedure:  Mix  the  powdered  galena  and  granulated  alumi- 
num and  place  the  mixture  in  the  clay  crucible  and  put  the  latter 
in  the  gas  furnace.  Place  the  cover  on  the  furnace  and  if  possible 
slip  it  a  little  toward  the  front,  so  that  by  standing  on  a  stool, 


126     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

one  can  look  down  through  the  hole  in  the  cover  and  see  the  charge 
in  the  crucible.  Light  the  furnace  and  heat  it  as  rapidly  as  pos- 
sible. Make  sure  the  iron  pan  is  dry  by  holding  it  with  the  tongs 
over  the  furnace.  Caution:  If  the  pan  should  be  wet,  the  heat 
of  the  molten  material  poured  into  it  would  cause  an  explosive 
formation  of  steam.  Watch  the  furnace  every  moment  of  the 
time  after  it  begins  to  get  hot.  A  white  puff  of  smoke  indicates 
that  the  reaction  is  taking  place.  If  one  can  see  the  charge 
one  can  see  an  incandescence  rapidly  spread  through  it.  After 
this  reaction,  leave  the  crucible  thirty  seconds  in  the  furnace, 
then  lift  it  with  the  tongs  and  pour  the  liquid  contents  into  the 
iron  pan.  Leave  the  latter  unmoved  until  the  contents  have 
solidified.  Then,  working  in  the  furnace  room,  crack  the  brittle 
aluminum  sulphide  from  the  lead;  place  the  latter  in  the  box  for 
scrap  lead,  and  pack  the  aluminum  sulphide  in  a  2-ounce  common 
bottle  fitted  tightly  with  a  cork  stopper.  Caution:  Aluminum 
sulphide  reacts  with  the  moisture  of  the  air  producing  hydrogen 
sulphide.  Do  not  take  any  part  of  the  preparation  outside  of 
the  furnace  room  except  the  well  stoppered  bottle  of  product. 
The  crucible  with  adhering  aluminum  sulphide  can  best  be  dis- 
posed of  by  putting  it  in  a  pail  of  water  as  soon  as  it  is  cool,  thus 
getting  rid  of  the  hydrogen  sulphide  all  at  once. 

QUESTIONS 

1.  Experiment.  Drop  a  small  lump  of  aluminum  sulphide 
(at  the  hood)  into  a  test  tube  of  water.  What  is  the  gas  formed 
and  what  is  the  insoluble  residue  left?  This  is  a  case  of  hy- 
drolysis. Write  the  equation  in  ionized  form,  and  name  the  acid 
and  the  base. 

s  2.  Lead  sulphide  does  not  hydrolyze  in  the  same  way.  Com- 
pare" the  basic  strength  of  Pb(OH)2  and  A1(OH)3.  Does  this 
fully  account  for  the  difference?  Perhaps  the  relative  solubility 
of  PbS  and  A12S3  would  account  for  some  of  the  difference. 
Explain  how.  Would  this  explanation  require  the  solubility 
of  A12S3  (as  such)  in  water  to  be  greater  or  less  than  that  of  PbS? 


CALCIUM  SULPHIDE  127 

PREPARATION  11 
CALCIUM  SULPHIDE 

This  preparation  is  to  be  made  by  the  reduction  of  the  oxy- 
salt  calcium  sulphate  by  means  of  carbon.  Calcium  is  too  ac- 
tive to  be  reduced  by  this  means,  but  the  sulphur  is  readily  re- 
duced from  its  valence  of  +  VI  to  its  minimum  valence  of  —  II. 

The  most  abundant  source  of  calcium  sulphate  is  the  mineral 
gypsum  CaS04.2H2O.  We  desire  anhydrous  CaSO4  as  our  start- 
ing material,  but  plaster  of  paris,  CaSO4.JH2O,  which  is  made  in 
enormous  quantities  by  gently  heating  gypsum,  is  so  available 
that  we  shall  choose  it  and  let  it  become  fully  dehydrated  in  the 
furnace. 

The  calcium  sulphide  prepared  by  the  following  procedure  is 
not  luminescent.  The  luminescent  varieties  owe  that  property 
io  traces  of  other  substances.  Directions  for  preparing  samples 
which  will  luminesce  with  almost  any  desired  color  are  given  in 
Vanino,  Handbuch  der  Praparativen  Chemie,  Volume  I,  p.  382. 

Materials:       plaster  of  Paris,  CaS04.iH2O,  145  grams  =  1  F.W. 
powdered  charcoal,  C,  48  grams  =  4  F.W. 

Apparatus:  t  clay  crucible  and  cover, 
gas  furnace. 

Procedure:  Unless  the  charcoal  is  already  very  finely  powd- 
ered, grind  it  thoroughly  in  a  large  porcelain  mortar.  Add  the 
plaster  of  Paris  and  mix  the  two  materials  and  pack  the  mixture 
in  the  clay  crucible.  Heat  the  crucible  to  between  a  bright 
red  and  a  yellow  heat  for  one  hour  and  a  half.  When  cold, 
inspect  the  contents  of  the  crucible,  particularly  the  inner  portions 
to  which  the  heat  would  have  penetrated  least;  there  should  be 
no  unburned  charcoal  left;  a  small  sample  should  dissolve  with 
effervescence  (hood)  in  hydrochloric  acid  and  leave  no  residue 
more  than  a  slight  turbidity.  If  the  reduction  has  not  been 
complete,  grind  the  material  again  after  adding  5  grams  of  char- 
coal, and  heat  it  again  for  an  'hour  and  a  half.  Preserve  the 
product  in  a  12-ounce  cork-stoppered  bottle. 

QUESTIONS  - 

1.  Write  the  equation  for  and  explain  the  behavior  of  calcium    \ 
sulphide  with  acids. 

2.  Treat  a  gram  of  calcium  sulphide  with  20  cc.  of  water. 


128      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

Is  there  any  visible  change?  Filter.  Add  hydrochloric  acid  to 
the  filtrate.  What  is  dissolved  in  the  filtrate?  What  is  the 
substance  on  the  filter? 

3.  Evidently  calcium  sulphide  hydrolyzes  extensively.  Ex- 
plain how  this  is  possible  without  the  evolution  of  hydrogen  sul- 
phide. Write  equation.  Compare  this  case  of  hydrolysis  with 
that  of  aluminum  sulphide. 

PREPARATION  12 
MERCURIC  SULPHIDE 

THis  substance  is  conveniently  made  by  direct  synthesis  from 
the  elements.  Two  modifications  of  mercuric  sulphide  are  known, 
one  black,  which  is  formed  first  in  this  preparation  and  also  by 
precipitation  of  mercuric  and  sulphide  ions,  and  the  other  a  bril- 
liant red  (vermilion),  which  is  more  stable  and  into  which  the 
black  form  tends  to  change. 

Materials:      mercury,  Hg,  20  grams  =  0.1  F.W. 

flour  of  sulphur,  8  grams. 

6-normal  KOH,  15  cc. 

saturated  Na2S03  solution  50  cc. 
Apparatus:     6-inch  evaporating  dish. 

large  porcelain  mortar. 

warm  closet  or  plate  at  about  50°. 

600  cc.  beaker. 

Procedure:  Place  the  mercury  and  the  sulphur  in  a  large 
mortar,  and  triturate  the  mixture,  moistening  it  with  ammonium 
sulphide  solution.  When  the  globules  of  metallic  mercury  seem 
to  have  entirely  disappeared,  add  the  KOH  solution  and  with  it 
rinse  as  much  as  possible  of  the  black  paste  into  the  6-inch  dish. 
Use  10  cc.  of  water  to  finish  rinsing  the  material  into  the  evapo- 
rating dish.  Let  this  dish  stand  in  a  place  at  a  temperature  of 
about  50°,  replacing  the  evaporated  water  as  often  as  each  day, 
and  stirring  the  mass  thoroughly  each  time  with  a  glass  spatula. 

When,  at  the  end  of  a  week  at  most,  the  mass  has  become  of  a 
pure  red  color,  wash  it  by  decantation  (see  Note  5(6),  page  10) 
in  a  tall  beaker  whereby  most  of  the  excess  of  sulphur  is  floated 
off.  Then  float  the  red  sulphide  itself  with  water  back  into  the 
evaporating  dish,  leaving  behind  in  the  beaker  any  lumps  of  black 


ALUMINUM   NITRIDE  129 

sulphide  or  globules  of  mercury.  Boil  the  red  sulphide  with  sodium 
sulphite  solution  to  remove  the  last  of  the  uncombined  sulphur; 
wash  by  decantation  with  boiling  water  and  collect  it  on  the 
suction  filter.  Dry  the  product  on  the  hot  plate  and  preserve  it 
in  a  cork-stoppered  test  tube. 

QUESTIONS 

1.  Red  and  black  sulphides  of  mercury  have  exactly  the  same 
composition,  as  expressed  by  the  empirical  formula  HgS.     Would 
you  regard  them  as  the  same  or  as  different  substances?     Give 
your  arguments. 

2.  What  explanation  can  you  give  of  the  action  of  sodium  hy- 
droxide in  facilitating  the  change  of  the  black  to  the  red  modi- 
fication? 

PREPARATION  13 
ALUMINUM  NITRIDE 

The  very  active  metals  are  capable  of  combining  directly  with 
nitrogen  to  form  nitrides.  In  the  air  the  oxide  is  formed  so  much 
more  readily,  that  nitride  formation  is  likely  to  escape  notice; 
but  if  the  metal  is  presented  in  powdered  form  in  a  thick  mass 
the  oxygen  is  all  combined  in  the  surface  layer  and  only  nitrogen 
penetrates  to  the  interior  where  pure  nitride  is  formed. 

Although  aluminum  is  a  very  active  metal,  it  enters  into  many 
reactions  with  extreme  difficulty  on  account  of  its  ability  to  cover 
itself  with  a  thin  tenaceous  coating  of  oxide,  which  keeps  it 
physically  separated  from  the  reacting  material.  Aluminum 
powder  alone  can  not  be  made  to  burn  in  air,  but  when  it  is  mixed 
with  lamp  black  and  any  part  is  brought  to  the  kindling  tem- 
perature, which  is  very  high,  the  combustion  spreads  throughout 
the  mass.  The  function  of  the  carbon  is  to  remove  the  surface 
film  of  oxide  from  the  grains  of  aluminum  powder,  and  the  action 
is  probably  as  follows:  It  is  well  known  that  carbon  cannot  re- 
duce aluminum  oxide  to  metal,  because  aluminum  is  more  ac- 
tive than  carbon.  But  carbon  and  aluminum  can  form  a  car- 
bide and  since  we  have  carbon  tending  to  take  oxygen  away  from 
aluminum  and  carbon  also  tending  to  take  aluminum  away  from 
oxygen  the  two  forces  working  together  bring  about  the  reaction 

2A1203  +  9C  -»  A14C3  +  6CO. 


130     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

In  an  atmosphere  of  nitrogen  however,  the  place  of  that  part  of 
the  carbon  which  combines  with  the  aluminum  may  be  taken 
by  nitrogen 

A1203  +  N2  +  3C  ->  2A1N  +  3CO. 

The  method  used  in  this  preparation  was  outlined  by  Doremus, 
Science  1920,  page  635. 

Materials:  finely  powdered  aluminum,  45  grams.  The  ma- 
terial sold  for  use  as  pigment  and  often  labeled 
Aluminum  Bronze,  although  nearly  pure  alu- 
minum, is  suitable  for  the  purpose.  The  oil 
which  still  adheres  from  the  grinding  is  of  no 
disadvantage. 

lamp  black,  5  grams. 

magnesium  ribbon,  4  inches. 

Apparatus:  8-inch  square  of  asbestos  paper  to  be  laid  on  a 
thick  iron  plate. 

large  porcelain  mortar. 

dark  colored  glasses  may  be  worn  to  protect  the 
eyes  from  the  blinding  light. 

Procedure:  Mix  the  aluminum  and  lamp  black  thoroughly 
in  a  mortar,  and  heap  up  the  mixture  in  as  compact  a  mound  as 
possible  on  the  asbestos  paper.  The  latter  must  be  placed  in  a 
safe  location  for  the  heat  will  be  so  intense  as  to  melt  through  the 
asbestos.  Insert  the  magnesium  ribbon  J  inch  into  top  of  pile 
and  light  the  free  end  from  the  gas  flame.  The  temperature 
of  the  burning  magnesium  is  high  enough  to  set  fire  to  the  mixture, 
but  the  spot  thus  ignited  is  quite  likely  to  cool  off  before  the  com- 
bustion can  get  well  started.  As  soon  as  the  ribbon  has  burned 
down  to  the  surface  of  the  pile,  play  the  gas  flame  over  the  hot 
spot,  until  the  combustion  is  thoroughly  under  way.  Although  the 
Bunsen  flame  alone  is  not  hot  enough  to  bring  the  mixture  to  the 
kindling  point,  it  prevents  the  spot  heated  by  the  magnesium  from 
cooling  so  rapidly.  Watch  the  combustion  spread  throughout 
the  mixture,  but  do  not  look  at  the  intense  light  for  more  than  a 
second  at  a  time  unless  the  eyes  are  protected  with  glasses.  When 
the  ash  is  cool,  break  it  up,  remove  as  much  of  the  aluminum  oxide 
crust  from  the  outside  as  may  be  and  preserve  the  nitride  in  a 
4-ounce  glass-stoppered  bottle. 


MAGNESIUM   NITRIDE  AND  AMMONIA  131 

QUESTIONS 

1.  Examine  the  aluminum  nitride  and  describe  its  appearance. 

2.  Treat  a  little  with  cold  and  hot  water.     Is  there  any  reaction? 

3.  Treat  a  little  with  NaOH  solution  and  warm.     Is  ammonia 
given  off?     What  type  of  reaction  is  this?     (See  next  preparation.) 

4.  Why  does  the  hydrolysis  of  aluminum  nitride  take  place  in 
NaOH  solution  but  not  in  pure  water?     (See  Exp.   9,   Chap. 

ID 

5.  Treat  portions  of  aluminum  nitride  with  6-normal  nitric, 
sulphuric,  and  hydrochloric  acid  respectively. 

6.  Heat  a  little  aluminum  nitride  in  the  open  air  to  see  how 
easily  it  can  be  converted  to  oxide. 

PREPARATION  14 
MAGNESIUM  NITRIDE  AND  AMMONIA 

The  formation  of  magnesium  nitride  follows  in  principle  the 
same  method  in  this  preparation  as  the  formation  of  aluminum 
nitride  in  the  preceding.  If,  however,  a  pile  of  magnesiuni  were 
allowed  to  burn  freely  in  the  open  air,  a  much  greater  conversion 
of  nitride  into  oxide  would  occur  before  the  reaction  died  out. 
Therefore  the  reaction  is  here  more  carefully  regulated  by  being 
carried  out  in  a  crucible  which  is  heated  from  an  external  source. 

Magnesium  nitride  differs  from  aluminum  nitride  in  that  it 
hydrolyzes  directly  with  pure  water;  in  fact  it  does  so  with  ex- 
treme violence. 

The  mechanism  of  the  hydrolysis  of  magnesium  nitride  is 
probably  similar  to  that  of  such  salts  as  sodium  carbonate  and 
ferric  chloride,  and  it  would  therefore  appear  as  follows : 


Mg3N2  -»  3Mg++ 
6H2O    -+60H- 

l 
Mg(OH)2  J, 

2N— 
6H+ 
I 
2NH3t 

If  magnesium  nitride  can  ionize  at  all  the  ions  which  it  would 
yield  are  obviously  Mg++  and  N  as  shown  in  the  upper  hori- 
zontal equation.  The  fact  that  such  an  ion  as  N~  ~  is  entirely 
unfamiliar  does  not  weaken  our  belief  in  the  above  mechanism, 
because,  as  the  direction  of  the  arrow  shows,  the  N ion  is 


132      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

formed  only  to  be  removed  according  to  the  right-hand  vertical 
equation. 

Materials:       powdered  magnesium,  10  grams. 

dry  sand,  50j;rams. 
Apparatus:     iron  crucible  and  cover  of  25  cc.  capacity. 

300  cc.  r.b.  flask. 

150  cc.  flask. 

filter  bottle  (or  500-cc.  flask). 

U-tube  J  inches  wide,  7|  inches  tall. 

2  thistle  tubes. 
1  pinch  cock. 

3  2-hole  rubber  stoppers. 
1  1-hole  rubber  stopper. 

See  diagram  for  fittings. 

Procedure:  Weigh  an  iron  crucible  of  about  25  cc.  volume  to- 
gether with  the  cover.  Pack  it  even  full  with  powdered  magne- 
sium, tapping  the  crucible  on  the  desk  to  make  the  powder  settle. 
Weigh  the  filled  crucible;  it  should  hold  about  10  grams  of  the 
powder.  Place  the  cover  tightly  on  top,  surround  the  crucible 
with  a  cylinder  of  asbestos  2J  to  3  inches  in  diameter  so  as  to 
diminish  the  loss  of  heat  by  radiation.  Heat  the  crucible  as  hot 
as  possible  with  a  Bunsen  burner  for  45  minutes.  After  it  cools 
empty  the  crucible  onto  a  piece  of  paper  and  note  the  white  MgO 
on  top  and  the  yellow  Mg3N2  beneath.  Place  the  material  in  a 
mortar,  break  up  the  lumps,  add  25  grams  of  dry  coarse  sand,  and 
mix  well.  Then  place  25  grams  of  sand  in  the  bottom  of  a  dry 
300-cc.  flask  and  pour  the  mixture  from  the  mortar  on  top  of  it. 
Use  this  as  the  generating  flask  A  in  the  diagram.  Have  the 
rest  of  the  apparatus  and  connections  fitted  as  in  the  diagram. 
Pour  50  cc.  of  water  into  the  absorption  flask  C.  The  water 
should  seal  the  bottom  of  the  thistle  tube  h  but  should  stand  about 
^  inch  below  the  end  of  the  delivery  tube  g.  Place  10  cc.  of  6-nor- 
mal  HC1  in  the  absorption  tube  F  and  then  add  enough  water 
to  seal  the  bend.  Remove  the  stopper  and  fittings  from  the 
generating  flask  A.  Pour  water  into  the  thistle  tube  i  and  open 
the  pinch  cock  until  the  stem  of  the  tube  has  filled  with  water. 
Replace  the  stopper  in  the  flask  and  open  the  pinch  cock  to  admit 
a  single  drop  of  water.  Add  another  drop  as  soon  as  the  reaction 
subsides  and  continue  to  add  a  single  drop  at  a  time  until  the 


MAGNESIUM   NITRIDE  AND  AMMONIA 


133 


reaction  becomes  less  violent.  Finally  add  enough  water  to 
make  70  cc.,  rock  the  flask  until  the  contents  are  thoroughly 
mixed,  then  while  still  rocking  the  flask  apply  a  small  flame  until 
the  liquid  boils.  Boil  gently  for  15  minutes.  Pour  together  the 


FIG.  18 

APPARATUS  FOR  DISTILLATION  OF   AMMONIA  FROM   MAGNESIUM   NITRIDE 

A  =  Generating  flask.  B  =  Trap  to  catch  solid  matter  entrained  with  gas 
and  steam.  C  =  Absorption  flask  with  pure  water  in  bottom.  D  =  Absorp- 
tion tube,  bend  sealed  with  dilute  acid. 

contents  of  the  absorption  flask  and  the  absorption  tube,  and, 
using  litmus  as  an  indicator,  add  enough  more  6-normal  HC1 
to  just  neutralize  the  ammonia.  Evaporate  the  solution  to  ob- 
tain solid  ammonium  chloride,  according  to  Note  6(6),  page  12. 
Preserve  the  product  in  a  6-inch  cork-stoppered  test  tube. 


134     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

QUESTIONS 

1.  Burn  a  little  calcium  in  the  air  and  test  the  ash  for  nitride. 
How? 

2.  Give  reasons  for  regarding  the  action  of  magnesium  nitride 
with  water  as  an  example  bf  hydrolysis.     Remember  that  hy- 
drolysis is  the  exact  reverse  of  neutralization  and  produces  an 
acid  and  a  base  from  a  salt  and  water.     What  is  the  acid  and  what 
is  the  base  in  this  case? 

3.  Why  is  it  necessary  to  mix  the  ^agnesium  nitride  with 
an  inert  material  such  as  sand  before  adding  water? 

4.  The  layer  from  the  top  of  the  crucible  will  often  contain  a 
black  substance  as  well  as  a  white,  and  particularly  so  if  the 
gases  from  the  flame  entered  under  the  lid  of  the  crucible.     What 
is  this  black  substance  and  why  does  it  form? 

5.  On  the  basis  that  air  contains  4  volumes  of  nitrogen  to  1 
volume  of  oxygen  and  that  all  the  oxygen  and  nitrogen  that  enter 
under  the  lid  of  the  crucible  combine  to  form  solid  oxide  and 
nitride,  figure  what  fraction  bf  the  magnesium  would  be  converted 
to  nitride.  .  , 

Experiments 
VALENCE 

The  elements  are  divided  into  two  classes,  metals  and  non- 
metals.  Chemically  the  properties  of  the  metals  are  as  different 
from  those  of  the  non-metals  as  physically,  for  the  metals  form  the 
electro-positive  constituents  of  compounds,  whereas  the  non-metals 
form  the  electro-negative  constituents. 

It  must  be  remembered  that  an  uncombined  element  is  elec- 
trically neutral,  but  every  element  has  a  more  or  less  strong  tend- 
ency to  assume  an  electrified  condition,  the  strength  of  which 
tendency  is  indicated  by  the  position  of  the  element  in  the  elec- 
tromotive series.  When  a  metal  and  a  non-metal  combine  chem- 
ically with  each  other  we  can  conceive  of  the  action  as  consisting 
merely  of  a  transfer  of  electricity  so  that  the  metallic  constituent 
becomes  positively  charged  and  the  non-metallic  negatively. 
The  electrostatic  attraction  between  the  charges  holds  the  con- 
stituents rigidly  in  place  in  a  solid  compound  and  so  the  solid  sub- 
stance is  a  non  conductor.  Pure  liquid  and  gaseous  compounds 
are  likewise  non  conductors,  the  electrostatic  forces  holding  the 


OXIDES  135 

constituents  of  each  individual  molecule  together.  But  we  have 
seen  in  Chapter  III  that  many  compounds,  viz.  acids,  bases  and 
salts,  give  a  conductive  solution  when  they  are  dissolved  in  water. 
The  compounds  are  ionized  in  the  solution.  Our  idea  of  ioniza- 
tion  by  no  means  signifies  that  the  opposite  charges  have  been 
separated  from  each  other;  it  simply  regards  the  single  positive 
constituent  as  no  longer  bound  exclusively  to  a  particular  negative 
constituent,  but  as  free  to  wander  about  through  the  solution 
holding  under  its  attractive  force  first  one  negative  constituent, 
then  another,  and  so  on. 

Were  it  not  for  the  discovery  of  ionization,  we  perhaps  would 
never  have  suspected  that  the  constituents  of  compounds  are 
electrically  charged.  That  this  is  the  case  however  seems  to  be 
a  well  warranted  conclusion,  and  it  is  for  this  reason  that  one 
speaks  of  metallic  elements  as  positive  and  non-metallic  elements 
as  negative  when  they  are  combined,  even  in  compounds  that 
are  not  ionized. 

The  number  of  charges  associated  with  a  sirhple  ion  as  A1+++  or 
Cl~  determines  its  valence,  the  valence  of  aluminum  thus  being 
+  III  and  that  of  chlorine  —  I  in  the  compound  aluminum  chlo- 
ride. Such  a  compound  of  two  elements  is  called  a  binary  com- 
pound and  the  constituents  are  held  by  the  forces  of  the  primary 
valence.  It  is  the  purpose  of  this  chapter  to  study  the  behavior 
of  the  non-metallic  elements  in  binary  compounds  in  which  they 
show  their  primary  negative  valence. 

OXIDES 

No  divalent  negative  ion  of  oxygen,  O  ,  has  ever  been  found; 
nevertheless  the  valence  of  oxygen  is  believed  to  be  —  II  because 
of  the  composition  of  water.  Nearly  all  of  the  elements  combine 
with  oxygen  forming  simple  binary  compounds. 

1.  Treat  a  little  cupric  oxide  with  hydrochloric  acid.  The 
liquid  acquires  a  blue  color  and  the  black  solid  finally  disap- 
pears entirely.  Evaporate  the  solution  and  a  residue  is  left 
which  becomes  brown  on  further  heating. 

The  brown  residue  is  anhydrous  cupric  chloride  in  which  the 
oxygen  of  cupric  oxide  has  been  exchanged  for  an  equivalent 
amount  of  chlorine. 

CuO  +  2HC1  ->  H20  +  CuCl2. 


136      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

The  primary  valence  of  all  of  the  elements  concerned  in  this  re- 
action has  remained  unchanged,  the  elements  have  simply  "  ex- 
changed partners,"  the  type  of  reaction  being  metathesis.  This 
instance  is  typical  of  the  reaction  of  metal  oxides  with  acids. 
The  oxygen  is  simply  exchanged  for  an  equivalent  amount  of  acid 
radical  and  the  valence  of  the  metal  is  not  altered. 

PEROXIDES 

Hydrogen  peroxide,  H202,  contains,  for  a  given  amount  of  hy- 
drogen, twice  as  much  oxygen  as  does  water  and  the  additional 
amount  of  oxygen  is  held  in  an  unstable  state  of  combination. 
Pure  hydrogen  peroxide  is  explosive;  in  dilute  solution  it  gives  off 
one-half  its  oxygen  more  or  less  rapidly  but  not  explosively.  In 
pure  solutions  this  decomposition  proceeds  very  slowly,  so  as  to 
be  practically  imperceptible;  but  many  substances  act  as  cata- 
lyzers for  the  reaction,  especially  solid  materials  which  have  a 
rough  surface,  notary  platinum  and  manganese  dioxide. 

2.  To  10  cc.  of  a  3  percent  solution  of  hydrogen  peroxide 
in  a  test  tube,  add  a  pinch  of  powdered  manganese  dioxide. 
Note  that  a  vigorous  effervescence  at  once  ensues  and  that 
the  escaping  gas  will  cause  a  glowing  splinter  to  burst  into 
flame. 

3.  Test  for  Hydrogen  Peroxide.     To  2  cc.  of  the  3  percent 
hydrogen  peroxide  solution  add  18  cc.  of  water,  thus  making 
the  volume  20  cc.     Mix  thoroughly,  pour  about  15  cc.  of  the 
solution  into  a  test  tube,  and  add  1  cc.  of  a  solution  of  titanium 
sulphate.*     This  is  one  of  the  standard  tests  for  hydrogen 
peroxide.     Determine  the  delicacy  of  the  test  by  repeating 
it  successively  with  smaller  and  smaller  amounts  of  hydro- 
gen peroxide,  as  follows:   take  2  cc.  of  the  remainder  of  the 
diluted  solution  and  dilute  it  to  20  cc.  by  adding  18  cc.  of 
water.     Test  15  cc.  of  this  solution  by  adding  1  cc.  of  titanium 
sulphate,  and  reserve  2  cc.  for  further  dilution.     Proceed  in 
this  way  with  successive  dilutions  according  to  the  powers 
of  ten  until  a  solution  is  obtained  which  shows  a  distinct 
color  with  the  titanium  sulphate,   while  the  next  dilution 

*  Prepared  by  fusing  1  part  of  titanium  dioxide  with  15  to  20  parts  of  po- 
tassium bisulphate  and  dissolving  the  mass  in  500  cc.  cold  dilute  sulphuric 
acid. 


OXIDES  137 

fails  to  show  the  test.  In  the  final  tests  add  1  cc.  of  reagent 
to  15  cc.  of  pure  water  and  hold  this  side  by  side  with  the 
sample  being  tested,  looking  through  the  length  of  the  column 
of  liquid  and  placing  a  piece  of  white  paper  for  a  background. 
By  comparing  with  a  "  blank  "  (the  pure  water  plus  reagent) 
in  this  way  the  fainter  colors  may  be  recognized  with  much 
more  certainty.  Express  the  sensitiveness  of  the  test  as  the 
number  of  parts  of  water  in  which  1  part  by  weight  of  hydro- 
gen peroxide  can  be  diluted  and  still  give  a  distinct  test. 

We  have  used  litmus  as  a  test  reagent  for  acids  and  bases 
although  we  probably  have  not  learned  the  chemical  formula  for 
litmus  nor  the  difference  in  nature  between  the  red  and  the  blue 
litmus.  When  we  find  by  experience  that  a  very  pronounced  effect 
is  always  shown  when  a  substance  is  treated  with  a  particular 
reagent,  we  can  use  this  reagent  as  a  test  reagent  for  the  substance. 
If  the  effect  is  given  the  presence  of  the  substance  is  proved.  In 
such  a  way  it  has  been  found  that  titanium  sulphate  gives  a  deep 
yellow  color  with  hydrogen  peroxide,  and  thus  we  use  this  reagent 
and  the  effect  as  a  test  for  hydrogen  peroxide,  without  worrying  as 
yet  too  much  about  the  chemical  formula  of  the  yellow  substance. 

We  are  now  going  to  investigate  the  behavior  of  a  number  of 
oxides  with  acids  and  we  are  going  to  use  the  titanium  sulphate 
reagent  to  discover  whether  or  not  hydrogen  peroxide  is  formed 
by  the  action. 

4.  Peroxides,  (a)  To  1  gram  of  barium  peroxide  BaO2 
in  a  small  beaker  add  15  cc.  of  water  and  then  6-normal 
HN03  drop  by  drop  with  continuous  stirring  until  the  solid 
has  dissolved.  Test  the  solution  for  hydrogen  peroxide  by 
adding  titanium  sulphate.  Treat  another  sample  of  barium 
peroxide  with  another  acid  in  order  to  convince  yourself 
whether  the  effect  could  have  been  caused  by  the  nitric  acid. 
Treat  1  gram  BaO2  with  5  cc.  of  water  and  5  cc.  6-normal 
H2S04.  Here  a  clear  solution  cannot  be  obtained  because 
insoluble  barium  sulphate  forms.  Filter  off  the  insoluble 
residue  and  test  the  filtrate  with  titanium  sulphate. 

(6)  Test  barium  oxide  BaO  in  the  same  way.  Since  this 
oxide  is  difficult  to  get  and  since  it  reacts  vigorously  with  water 
anyway  to  form  the  hydroxide  Ba(OH)2  we  might  as  well  use 


138     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

the  hydroxide  as  a  starting  point,  but  should  bear  in  mind 
that  it  is  essentially  the  same  thing  as  the  oxide  BaO. 

(c)  Heat  a  piece  of  broken  porcelain  on  a  triangle  to  red- 
ness; drop  upon  it  a  piece  of  sodium  metal  the  size  of  a  small 
pea,  and  remove  the  flame.     The  sodium  burns  to  form  sodium 
peroxide  Na202.     Place  50  cc.  of  cold  water  and  5  cc.  of  6- 
normal  H2S04  in  a  small  dish;  let  the  porcelain  and  adhering 
sodium  peroxide  cool  completely  and  then  drop  it  into  the 
dilute  acid.     Test  the  solution  with  titanium  sulphate. 

(d)  Prepare  another   sample  of  sodium  peroxide   in  the 
same  way.     Scrape  it  off  the  porcelain  and  into  a  dry  test 
tube.     Add  5  cc.  of  water.     Note  the  effervescence  and  that 
the  gas  inflames  a  glowing  splinter.     Boil  the  solution  until 
effervescence  ceases.     Acidify  with  sulphuric  acid  and  test 
with  titanium  sulphate. 

In  (d)  an  amount  of  oxygen  is  evolved  corresponding  to  one-half 
the  oxygen  of  the  Na2O2.  (See  also  discussion  of  Experiment  11 
in  Chapter  II.)  This  leaves  in  the  solution  the  oxide  Na2O  or 
rather  NaOH  the  product  of  the  reaction  of  Na^O  with  water. 
On  acidification  we  get  a  simple  neutralization: 

2NaOH  +  H2SO4  ->  NaaSO,  +  2H2O, 
or,  Na2O  +  H2SO4  -»  NasSC^  +  H2O. 

But  in  (c)  the  peroxide  has  not  been  given  a  chance  to  decompose 
before  the  treatment  with  acid.  Since  hydrogen  peroxide  was 
found  as  a  product,  the  reaction  is  probably 

Na2O2  +  H2SO4  -»  Na2SO4  +  H202. 

Evaporation  of  the  solution  does  yield  crystals  of  Na2S04.10H2O 
identical  in  every  respect  to  the  sodium  sulphate  that  can  be 
obtained  from  (d)  thus  confirming  the  above  equations. 

In  the  same  way  barium  oxide  BaO  neutralizes  an  acid,  with  the 
formation  of  water: 

BaO  +  2HNO3  **  Ba(NO3)2  +  H2O 
and  barium  peroxide  yields  hydrogen  peroxide: 

BaO2  +  2HN03  ->  Ba(N03)2  +  H202. 

It  is  obvious  that  the  reaction  of  the  peroxide  with  an  acid  is  a 
metathesis  in  which  the  O2  radical  is  concerned  just  as  the  O  is 


OXIDES  139 

concerned  in  the  neutralization  of  an  ordinary  oxide.  It  is  fur- 
thermore obvious  from  the  formulas  H2O2  and  Na202  and  BaO2, 
if  we  ascribe  the  ordinary  valence  to  hydrogen,  sodium  and  barium 
that  the  valence  of  the  O2  radical  is  two. 

Thus  the  peroxides  are  not  really  simple  binary  compounds  in 
which  the  primary  positive  valence  of  the  metallic  constituent  is 
just  equalled  by  the  primary  negative  valence  of  the  non-metallic 
constituent.  The  peroxides  could  rather  be  classed  with  such 
compounds  as  nitrates  and  sulphates  in  which  there  is  a  complex 
negative  radical. 

5.  Dioxides.     To  about  a  gram  each  of  manganese  diox- 
ide, MnO2,  and  lead  dioxide,  PbO2,  in  separate  test  tubes  add 
10  cc.  of  water  and  5  cc.  of  6-normal  H2SO4  or  HN03  and 
warm  for  a  few  moments.     Note  that  the  dark  colored  pow- 
ders do  not  dissolve  or  change  their  appearance  in  any  way 
nor  is  there  any  effervescence.     Filter  off  the  insoluble  pow- 
der and  add  titanium  sulphate  to  the  filtrate.     Note  that 
there  is  no  yellow  coloration. 

These  two  oxides  do  not  react  at  all  with  these  acids,  and  indeed 
we  should  not  expect  on  that  account  to  find  any  hydrogen  perox- 
ide produced.  These  results  are  merely  negative  therefore  and 
leave  us  in  doubt  as  to  the  nature  of  the  oxides,  whether  they  are 
ordinary  binary  oxides  in  which  case  the  valence  of  the  manganese 
or  lead  is  four,  or  whether  they  are  peroxides  in  which  the  valence 
of  the  metal  is  two  like  that  of  barium  in  barium  peroxide. 

6.  Different  Behavior  of  Dioxides  and  Peroxides  with  Hy- 
drochloric Acid.     Treat  J  gram  of  finely  powdered  lead  dioxide 
with  5  cc.  of  ice  cold  12-nHCl.     Note  that  a  yellow  solution 
is  formed.     Dilute  1  cc.  of  the  solution  with   100  cc.  water 
and  add  1  cc.  of  titanium  sulphate.     Note  that  no  yellow 
color  is  produced.     Heat  the  rest  of  the  solution.     Note  that 
an  evil  smelling  yellow  gas  (chlorine)  is  given  off,  that  the 
yellow  color  disappears  and  that  a  white  crystalline  powder 
settles  out  as  the  solution  cools.     Treat  J  gram  of  barium 
peroxide  in  the  same  way  with  5  cc.  of  ice-cold  12-nHCl. 
Add  cold  water  until  the  crystalline  residue  is  all  dissolved; 
dilute  5  cc.  of  this  solution  with  100  cc.  of  water  and  add  1 
cc.  of  titanium  sulphate.     The  yellow  color  indicating  the 


140     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

presence  of  hydrogen  peroxide  is  obtained.  Warm  the  re- 
mainder of  the  solution.  Note  that  little  or  no  chlorine  is 
evolved.  Dilute  5  cc.  of  the  remaining  solution  with  100  cc. 
of  water  and  again  add  titanium  sulphate.  Note  that  the 
test  for  hydrogen  peroxidg  is  still  obtained  unless  the  solu- 
tion had  been  heated  too  long. 

The  yellow  soluble  substance  is  lead  tetrachloride  PbCl4  and 
is  formed  by  the  reaction 

Pb02  +  4HC1  -»  PbCl4  +  2H20. 

Lead  tetrachloride  is  very  unstable  and  decomposes  rapidly  into 
the  lower  chloride  PbCl2  and  free  chlorine.  This  chloride  of  lead 
is  the  same  one  that  is  obtained  when  lead  monoxide  is  treated 
with  hydrochloric  acid: 

PbO  +  2HC1  ->  PbCl2  +  H2O. 

Since  the  2  oxygens  of  lead  dioxide  are  exchangeable  for  4  chlo- 
rines, and  the  lead  is  exchangeable  for  4  hydrogens  of  the  acid, 
the  valence  of  lead  is  established  as  four  and  therefore  each  of  the 
divalent  oxygens  are  held  as  separate  units  by  the  lead. 

The  barium  peroxide  acts  as  we  should  expect  with  hydrochloric 
acid  yielding  hydrogen  peroxide.  Since  hydrogen  peroxide  is 
unstable,  breaking  down  into  water  and  oxygen  we  would  expect 
that  it  would  react  with  the  excess  of  hydrochloric  acid 

H2O2  +  2HC1  ->  2H2O  +  C12 

and  set  free  chlorine. 

This  experiment  therefore  has  shown  that  lead  dioxide  is  an 
ordinary  oxide.  The  prefix  di  indicates  the  quantity  of  oxygen 
and  that  the  valence  of  the  metal  is  sufficient  to  hold  all  of  the 
oxygen  in  the  ordinary  manner. 

THE  HALOGENS 

Recall  that  chlorine  is  a  greenish  yellow  gas,  bromine  is  a  dark 
red  liquid  which  readily  vaporizes  to  a  red  gas,  and  iodine  is  a 
nearly  black  solid  which  is  changed  by  heat  to  a  beautiful  violet- 
colored  gas. 

7.  Test  for  the  Presence  of  Iodine.    Volatility  of  Iodine. 
Place  some  small  grains  of  iodine  in  the  bottom  of  a  2-liter 


THE  HALOGENS  141 

bottle  and  suspend  a  piece  of  filter  paper  moistened  with 
starch  paste  in  the  upper  part  of  the  bottle.  Cover  the 
whole  with  a  watch  glass,  allow  it  to  stand  15  minutes  or  more 
and  observe  that  the  paper  slowly  turns  blue. 

This  experiment  not  only  shows  the  volatility  of  iodine, —  it  must 
pass  through  the  space  between  the  solid  grains  and  the  paper  as 
a  vapor  — ,  but  it  illustrates  one  of  the  striking  properties  of  free 
iodine,  namely  its  power  of  turning  starch  deep  blue.  We  will 
not  concern  ourselves  as  to  what  the  blue  substance  is,  but  we  shall 
employ  starch  in  testing  for  iodine,  and  look  for  the  blue  color  to 
indicate  its  presence. 

8.  Iodide-Starch  Paper.     Add  1  cc.  of  potassium  iodide 
solution  to  10  cc.  of  a  starch  solution.     Wet  a  number  of 
strips  of  filter  paper  with  the  solution,  allow  them  to  dry,  and 
save  for  use  in  certain  of  the  following  experiments.     The 
test  papers  thus  prepared  are  colorless. 

Free  iodine  colors  starch  blue,  but  iodine  in  combination  has 
altogether  different  properties. 

9.  Test  for  the  Presence  of  Chlorine  or  Bromine.     Place 
5  cc.  each  of  chlorine  and  bromine  water  in  separate  wide- 
mouthed  bottles.     Lower  for  a  moment  strips  of  moistened 
iodide-starch  paper  in  the  mouth  of  each  bottle.     The  papers 
are  immediately  turned  deep  blue. 

Chlorine  and  bromine  are  higher  in  the  electromotive  series 
than  iodine  and  thus  are  able  to  drive  it  out  of  the  ionic  form 

C12  +  2I-->2C1-  +  I2. 

10.  Chlorine  from  Hydrochloric  Acid.     Place  about  0.5 
gram  each  of  manganese  dioxide,  lead  dioxide,  potassium  or 
sodium  dichromate,  and  potassium  permanganate  in  separate 
test  tubes.     Add  about  2  cc.  of  6-normal  hydrochloric  acid 
to  each  and  test  for  chlorine  by  holding  iodide-starch  paper 
in  the  mouths  of   the  tubes.     Also  after  warming  a  very 
little  observe  the  odor  and  color  of  the  gas.     Rinse  out  the 
tubes  immediately  at  the  hood  under  the  sink.     Compare 
the  action  of  the  oxides  used  above  with  that  of  copper  oxide 
CuO  and  lead  oxide  PbO. 


142      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

The  action  of  the  lower  oxides  is  one  of  simple  metathesis, 
chlorine  and  oxygen  simply  exchange  places,  no  chlorine  being 
set  free. 

CuO  +  2HC1  -*  CuCl2  +  H2O. 

There  are  several  ways  of  explaining  the  mechanism  of  the 
reaction  in  which  chlorine  is  liberated.  Let  us  imagine  that  the 
first  step  is  a  metathesis: 

MnO2  +  4HC1  ->  MnCl4  +  '2H2O. 
But  manganese  tetrachloride  is  not  stable  and  it  decomposes 

MnCl4-^MnCl2  +  C12; 
adding  these  equations  we  obtain 

Mn02  +  4HC1  ~>  MnCl2  +  C12  +  2H2O. 

This  is  a  reaction  of  oxidation  and  reduction  (see  page  105);  one 
atom  of  manganese  changes  its  valence  from  +  IV  to  +  II  and 
this  is  compensated  by  the  change  of  two  atoms  of  chlorine  from 
—  I  to  zero,  so  that  algebraically  the  total  changes  of  valence  add 
to  zero. 

The  reaction  of  lead  dioxide  may  be  similarly  figured  out  if  one 
has  the  information  that  the  stable  chloride  of  lead  is  PbCl2 
corresponding  to  the  oxide  PbO. 

The  reactions  of  potassium  dichromate  and  potassium  per- 
manganate are  represented  by  the  equations 

K2Cr207  +  14HC1  -*2KC1  +  2CrCl3  +  3C12  +  7H20 
2KMnO4  +  16HC1  -*  2KC1  +  2MnCl2  +  5C12  +  8H20. 

These  are  likewise  reactions  of  oxidation  and  reduction  and  the 
valence  changes  add  up  as  follows: 

2Cr      +  VI  to  +  III  2  X  (-  3)  =  -  6 

6C1       -  I     to        0  6  X  (+  1)  =  +  6 


Total  change  =  0 

2Mn    +VII  to  +  II  2  X  (-  5)  =  -  10 

10C1     -  I       to      0  10  X  (+  1)  =  +  10 

Total  change  =  0 

In  working  out  these  equations  it  would  be  necessary  to  know 
that  the  stable  salt  forming  oxides  of  chromium  and  manganese  are 


THE  HALOGENS  143 

C^Os  and  MnO,  corresponding  to  the  chlorides  CrCls  and  MnCl2 
respectively. 

Let  us  analyze  the  possible  mechanism  of  one  of  these  reactions : 
Potassium  permanganate  is  an  oxy-salt  which  could  be  formed 
from  a  basic  oxide  and  an  acid  oxide  (see  page  61).  Let  us 
imagine  the  first  step  of  the  reaction  is  to  resolve  this  salt  into  its 
simple  oxides. 

2KMn04  -»  K2O  +  Mn207 

Potassium  oxide,  being  a  strongly  basic  oxide  would  react  with 
the  acid  without  any  question 

K2O  +  2HC1  -»  2KC1  +  H2O 

and  manganese  heptoxide  could  be  imagined  to  undergo  a  meta- 
thesis with  hydrochloric  acid 

Mn2O7  +  14HC1  -»  2MnCl7  +  7H20 

but  the  existence  of  the  heptachloride  is  only  hypothetical  since 
such  a  substance  has  never  been  isolated;  it  will  therefore  be 
expected  to  decompose  as  soon  as  formed: 

2MnCl7  ->  2MnCl2  +  5C12. 

Now  these  separate  steps  may  be  added  together  and  give  identic- 
ally the  same  equation  as  printed  above  for  the  complete  reaction. 

We  may  summarize  this  experiment  by  the  statement  that 
free  chlorine  is  liberated  from  hydrochloric  acid  by  strong  oxidiz- 
ing agents.  Whether  or  not  the  oxidizing  agent  is  strong  enough 
to  do  this  may  perhaps  be  foretold  by  considering  the  element 
which  has  the  higher  than  ordinary  valence ;  if  the  chloride  of  this 
element  in  which  the  higher  valence  would  be  satisfied  is  unstable, 
then  the  oxidizing  agent  will  set  chlorine  free. 

11.  Bromine   and   Iodine   from   Bromides   and   Iodides. 

Test  the  action  of  any  one  of  the  oxidizing  agents  used  in 
No.  10,  say  manganese  dioxide,  on  hydrobromic  and  hydri- 
odic  acids. 

Add  a  few  drops  of  chlorine  water  to  5  cc.  of  a  bromide 
solution,  for  example  NaBr.  Then,  in  order  to  find  whether 
bromine  has  been  set  free,  add  1  cc.  of  carbon  disulphide, 
shake  vigorously,  and  let  the  heavier  liquid  settle  to  the 
bottom.  The  free  halogen  is  more  soluble  in  carbon  di- 
sulphide than  in  water,  consequently  it  dissolves  in  and  im- 


144      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

parts  its  characteristic  color  to  it.     Note  that  the  globule 
has  acquired  an  orange  red  color. 

Likewise  add  a  few  drops  of  chlorine  water  and  of  bro- 
mine water  to  separate  portions  of  an  iodide  solution,  and 
test  in  each  case  with  carbon  disulphide.  Note  that  in  each 
case  the  globule  becomes  violet  colored. 

Although  this  experiment  is  almost  a  repetition  of  No.  9,  it 
does  illustrate  another  method  of  recognizmg  small  amounts  of 
free  bromine  and  iodine.  It  emphasizes  again  that  corresponding 
to  their  position  in  the  electromotive  series,  chlorine  is  more  ac- 
tive than  bromine  and  bromine  in  turn  more  active  than  iodine. 

Iodine  thus  can  be  liberated  by  the  weakest  oxidizing  agents. 
It  should  be  noted  that  in  this  connection  the  halogens  themselves 
are  oxidizing  agents;  for,  in  the  reaction 


2KI  +  Br2  ->  2KBr  +  I 


the  bromine  has  gone  from  a  valence  of  zero  to  —I,  that  is,  it 
has  been  reduced,  while  the  iodine  has  been  oxidized  from  val- 
ence —  I  to  valence  zero. 

Fluorine  is  the  strongest  oxidizing  agent  of  all.  It  liberates 
the  other  halogens  from  their  compounds.  It  will  even  liberate 
oxygen  from  water  and  it  is  of  particular  interest  that  a  portion 
of  the  oxygen  so  liberated  is  in  the  more  active  form  of  ozone.  The 
oxidizing  power  of  fluorine  must  be  extraordinarily  high  in  order 
to  build  up  the  very  active  oxidizing  agent  ozone.  It  is  impossible 
to  liberate  fluorine  from  hydrofluoric  acid  or  fluorides  by  any  of 
the  chemical  oxidizing  agents  because  there  is  no  other  electro- 
negative element  which  exceeds  it  in  activity.  It  is  impossible 
to  liberate  it  from  an  aqueous  solution  by  electrolysis  because  the 
less  active  oxygen  is  set  free  instead.  But  fluorine  can  be  pre- 
pared by  electrolysis  of  a  solution  of  potassium  fluoride  in  an- 
hydrous hydrogen  fluoride.  This  solution  is  a  good  electrolyte 
and  since  it  contains  no  other  negative  ion  than  fluoride,  it  is 
fluorine  that  has  to  be  discharged  at  the  anode. 

FORMATION  OF  AND  PROPERTIES  OF  THE  HYDROGEN  HALIDES 

12.  Hydrogen  Chloride.  Add  2  cc.  of  concentrated  sul- 
phuric acid  (36-normal)  to  about  0.5  gram  of  sodium  chloride 
in  a  test  tube  and  warm  it  a  very  little  if  necessary.  Test  for 


HYDROGEN  HALIDES  145 

hydrogen  chloride  gas  by  blowing  gently  across  the  mouth 
of  the  tube,  by  holding  moistened  litmus  in  the  gas,  and  by 
bringing  a  strip  of  filter  paper  moistened  with  ammonia 
water  near  the  mouth  of  the  tube.  The  salt  effervesces  in 
the  concentrated  acid.  The  gas  issuing  from  the  tube 
creates  a  dense  fog  with  the  breath.  The  gas  turns  moistened 
litmus  red  and  it  produces  a  dense  white  smoke  in  the  vicinity 
of  the  paper  moistened  with  ammonia  water. 

The  main  reaction  consists  of  the  displacement  of  a  volatile 
acid  from  its  neutral  salt  by  means  of  a  non  volatile  acid. 

NaCl  +  H2SO4  ->  NaHSO4  +  HC1 1. 

Although  hydrogen  chloride  is  excessively  soluble  in  water  it  is 
not  soluble  in  sulphuric  acid. 

The  fogging  of  the  breath  is  caused  by  a  condensation  of  the 
water  vapor.  This  condensation  is  possible  because  of  the  ex- 
treme solubility  of  hydrogen  chloride;  such  a  concentrated  solu- 
tion can  be  formed  that  the  tension  of  the  water  vapor  which  can 
escape  from  it  is  reduced  to  a  very  low  point.  The  water  vapor 
in  the  breath  is  at  a  much  higher  pressure  and  thus  being  out  of 
equilibrium  it  condenses.  The  fog  then  consists  of  countless 
minute  globules  of  hydrochloric  acid  solution. 

The  hydrogen  chloride  of  course  dissolves  in  the  water  in  the 
moist  litmus  paper  and  makes  hydrochloric  acid. 

The  smoke  with  ammonia  paper  is  due  to  the  precipitation  of 
solid  ammonium  chloride  where  the  gases  HC1  and  NH3  meet. 
The  ammonium  hydroxide  dissociates  non-electrolytically 

NH4OH  ^±  NH3  +  H20 

and  the  ammonia  combines  with  the  hydrochloric  acid 
NH3  +  HC1  -» NH4C1. 

These  properties  of  fogging  the  breath,  reddening  litmus,  and 
making  a  smoke  with  ammonia,  are  characteristic  of  hydrogen 
chloride  and  the  other  hydrogen  halides,  and  one  can  infer  that  a 
hydrogen  halide  is  present  when  these  properties  are  observed. 

13.  Hydrogen  Bromide.     Add   1   cc.    of    36-normal    sul- 
phuric acid  to  about  J  gram  of  powdered  potassium  or  so- 


146      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

dium  bromide  in  a  test  tube.  Apply  the  same  tests  as  in 
No.  12  then  look  for  other  new  substances  formed,  apply- 
ing the  following  tests,  and  towards  the  last  heating  the  tube 
a  little.  Observe  the  color  of  the  gas  and  also  the  effect  of 
lowering  iodide-starch  papier  for  a  moment  only  into  the  tube. 
Continued  exposure  of  iodide-starch  paper  to  strong  acid 
fumes  will  develop  a  blue  color  in  any  case.  Observe  the 
odor,  but  with  great  caution.  Test  the  gases  with  a  strip 
of  filter  paper  moistened  with  lead  acetate  solution.  The 
salt  effervesces  in  the  concentrated  acid.  The  gas  evolved 
fogs  the  breath  even  more  strongly  than  hydrogen  chloride. 
It  reddens  litmus  and  it  gives  a  dense  smoke  with  ammonia. 
The  gas  is  quite  strongly  tinged  with  red;  it  turns  iodide- 
starch  paper  blue  immediately;  it  does  not  darken  lead 
acetate  paper.  Sometimes  one  is  able  to  distinguish  the 
odor  of  sulphur  dioxide,  even  in  the  presence  of  all  the  other 
smells. 

The  fogging,  litmus  test,  and  smoking  with  ammonia  indicate 
that  hydrogen  bromide  is  freely  evolved.  The  reddish  tinge  to 
the  gas  and  the  coloring  of  iodide-starch  paper  indicate  the  pres- 
ence of  a  rather  small  amount  of  free  bromine.  The  failure  to 
darken  lead  acetate  indicates  absence  of  hydrogen  sulphide. 

The  principal  reaction  in  this  experiment  is  similar  to  that  in 
the  preceding  experiment: 

NaBr  +  H2SO4  ->  NaHSO4  +  HBr 

The  properties  of  hydrogen  bromide  are  very  similar  to  those  of 
hydrogen  chloride  but  differ  in  that  hydrogen  bromide  reacts  with 
sulphuric  acid 

H2S04  +  2HBr  -*  H2SO3  +  H2O  +  Br2 
H2S03  -*  H2O  +  SO2 

which  reactions  account  for  the  red  color  of  free  bromine,  the 
turning  blue  of  iodide-starch  paper,  and  the  odor  of  sulphur  diox- 
ide. The  major  part  of  the  hydrogen  bromide,  it  is  true,  escapes 
from  the  reaction  mixture  unaffected,  but  on  bubbling  up  through 
the  concentrated  sulphuric  acid  a  small  part  of  it  is  oxidized  ac- 
cording to  the  secondary  reaction. 
It  is  recalled  from  Experiment  11,  that  bromine  is  a  less  active 


HYDROGEN  HALIDES  147 

element  than  chlorine;  hence  it  is  but  natural  that  we  should 
find  it  here  displaced  from  its  hydrogen  compound  by  a  weaker 
oxidizing  agent.  Chlorine  is  displaced  from  hydrogen  chloride 
by  strong  oxidizing  agents,  Mn02,  Pb02,  K2Cr2O7,  KMnO4,  but 
bromine  is  displaced  by  the  comparatively  weak  oxidizing  agent 
H2SO4  which  has  no  action  on  hydrogen  chloride.  The  sulphur 
of  the  sulphuric  acid  is  reduced  from  valence  +  VI  to  valence 
+  IV  in  sulphur  dioxide. 

14.  Hydrogen  Iodide.  Add  1  cc.  of  36-normal  sulphuric 
acid  to  \  gram  of  powdered  potassium  or  sodium  iodide, 
and  apply  all  of  the  test  enumerated  in  Nos.  12  and  13; 
also  inspect  the  walls  of  the  test  tube  carefully  to  see  if  any 
solid  sulphur  condenses. 

As  in  Nos.  12  and  13  the  solid  effervesces  in  the  concen- 
trated sulphuric  acid  and  the  gas  evolved  fogs  the  breath 
(even  more  markedly  in  this  case)  turns  litmus  red,  and  gives 
a  dense  smoke  with  ammonia.  When  the  tube  is  warmed, 
the  beautiful  purple  iodine  vapor  is  seen  inside  and  nearly 
black  crystals  collect  on  the  cooler  upper  walls. 

Lead  acetate  paper  is  colored  dark  brown  and  the  odor  of 
hydrogen  sulphide  is  very  obvious.  Sometimes  a  powdery 
light  yellow  substance  (sulphur)  is  seen  collecting  on  the 
walls  of  the  tube. 

The  primary  reaction  in  this  experiment, 

Nal  +  H2S04  -» NaHS04  +  HI, 

is  of  the  same  nature  as  that  in  the  preceding  two  experiments. 
That  hydrogen  iodide  escapes  abundantly  from  the  reaction  mix- 
ture is  attested  by  the  fogging  of  the  breath,  by  the  reddening  of 
litmus  and  by  smoke  with  ammonia.  But  that  hydrogen  iodide 
enters  more  extensively  into  a  secondary  reaction  with  sulphuric 
acid  is  shown  by  the  abundance  of  the  secondary  products.  These 
consist  of  free  iodine,  shown  by  the  purple  vapor  and  the  black 
crystalline  deposit;  hydrogen  sulphide,  shown  by  the  darkening 
of  lead  acetate  paper  (PbAc2  +  H2S  — »  PbS  +  2HAc),  and  the 
odor;  sulphur,  shown  by  the  light  yellow  deposit.  Their  forma- 
tion can  be  accounted  for  by  the  equations: 

H2SO4  +  SHI  ->  H2S  +  4H2O  +  4I2. 


148      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

Hydrogen  sulphide  and  iodine  do  not  react  in  the  gaseous  condi- 
tion but  where  they  dissolve  in  the  film  of  moisture  condensed  on 
the  cooler  walls  of  the  tube  they  react 

H2S  +  I2  ->  2HI  +  S 

and  this  accounts  for  the  deposit  of  sulphur. 

This  experiment,  taken  in  conjunction  with  the  preceding  two 
shows  the  gradation  in  properties  of  the  hydrogen  halides.  The 
activity  of  the  halogens  decreases  in  the  order:  fluorine,  chlorine, 
bromine,  iodine.  Hydrogen  fluoride  is  so  stable  therefore  that  no 
chemical  substance  can  displace  fluorine,  and  hydrogen  fluoride 
is  unable  to  act  as  a  reducing  agent  in  any  circumstance.  Hy- 
drogen chloride  can  act  as  a  reducing  agent  upon  the  most  powerful 
oxidizing  agents.  Hydrogen  bromide  can  reduce  the  compara- 
tively weak  oxidizing  agent  sulphuric  acid,  but  it  can  only  carry 
the  valence  of  the  sulphur  down  to  +  IV  in  sulphur  dioxide.  Hy- 
drogen iodide  can  reduce  sulphuric  acid  much  more  freely  carrying 
the  valence  all  the  way  from  +  VI  down  to  the  lowest  possible, 
namely  —  II  in  hydrogen  sulphide. 

15.  Hydrogen  bromide  and  hydrogen  iodide  might  be 
obtained  pure  from  their  salts,  provided  the  latter  were 
treated  with  a  non-volatile  acid,  such  as  sulphuric  acid, 
but  one  which  would  not  at  the  same  time  behave  as  an  ox- 
idizing agent. 

Place  about  J  gram  of  powdered  sodium  bromide  in  a  test 
tube  with  a  little  concentrated  phosphoric  acid  solution,  heat, 
and  test  the  evolved  gas  for  free  halogen  as  well  as  for  hy- 
drogen bromide.  Repeat,  using  sodium  iodide  in  place  of 
the  sodium  bromide. 

In  both  experiments,  fogging  of  the  breath,  reddening  of 
litmus,  and  smoke  with  ammonia  are  noted.  With  the  so- 
dium bromide,  the  gas  is  entirely  uncolored  and  no  test  with 
iodide  starch  paper  is  given.  With  sodium  iodide  the  gas 
is  at  first  entirely  colorless  but  when  the  tube  is  heated 
strongly  the  film  of  moisture  condensing  on  the  walls  of  the 
tube  becomes  slightly  brown.  No  test  is  given  with  starch 
paper. 

Phosphoric  acid  is  a  much  weaker  oxidizing  agent  than  sul- 
phuric acid  and  it  fails  to  oxidize  either  hydrogen  bromide  or 


REACTIONS  OF  HALIDE  IONS  149 

hydrogen  iodide.  The  brown  color,  it  is  true,  indicates  a  trace 
of  free  iodine  but  this  is  accounted  for  by  a  direct  decomposition 
of  hydrogen  iodide  by  heat. 

CHARACTERISTIC  REACTIONS  OF  THE  HALIDE  IONS 

The  halides  of  all  the  metals  except  silver,  mercury,  and  lead 
are  soluble  in  water,  but  with  the  ions  of  these  three  metals,  the 
halide  ions  for  the  most  part  give  characteristic  precipitates. 
The  precipitates  are  valuable  as  tests  for  identifying  either  the 
halogens  or  the  metals  in  qualitative  analysis. 

16.  (a)  Add  a  few  drops  of  silver  nitrate  solution  to  a 
few  drops  each  of  hydrochloric,  hydrobromic,  and  hydriodic 
acid  solutions,  diluted  with  5  cc.  of  water,  in  separate  test 
tubes.  Let  each  precipitate  settle,  pour  off  most  of  the 
liquid,  and  find  if  portions  of  the  precipitate  dissolve  in  a 
large  amount  of  boiling  water.  Test  also  the  solubility  of 
each  in  ammonia. 

(6)  Add  lead  nitrate  solution  also  to  each  of  the  three 
acids  and  test  the  solubility  of  the  precipitates  in  hot  water. 

Tabulate  the  results  obtained  in  (a)  and  (6). 

The  exact  figures  for  the  solubility  at  different  temperatures 
may  be  looked  up  in  the  solubility  table  in  the  Appendix.  The 
solubility  of  the  silver  halides  in  ammonia  solution  depends  on  the 
formation  of  the  complex  ion  (Ag.2NH3)  +  (see  chapter  III,  page 
97).  Silver  bromide  and  silver  iodide  are  decreasingly  less  sol- 
uble in  pure  water  than  the  chloride,  although  all  three  seem  by  the 
direct  experiment  to  be  completely  insoluble.  Thus  in  a  solution 
containing  ammonia,  the  (Ag.2NH3)+  ion  does  not  dissociate 
enough  for  the  simple  silver  ions  and  chlorine  ions  to  reach  the 
solubility  product  of  silver  chloride.  The  solubility  product  of 
silver  bromide  is  sooner  reached  and  silver  bromide  is  only  sparingly 
soluble  in  ammonia  solution.  The  solubility  product  of  silver 
iodide  is  still  smaller  and  this  accounts  for  the  apparent  complete 
insolubility  of  silver  iodide  in  ammonia  solution. 

RELATIVE  ACTIVITY  OF  THE  HALOGENS,  OXYGEN,  AND  SULPHUR 

The  activity  of  an  element  is  judged  by  its  power  to  combine 
with  other  elements,  or  by  its  ability  to  pass  from  the  elementary 
condition  into  solution  in  the  form  of  ions,  whereby  it  forces  other 


150     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

less  active  elements  out  of  the  ionic  into  the  elementary  condition. 
There  can  be  no  doubt  as  to  the  relative  activity  of  the  halogens 
as  regards  their  tendency  to  form  simple  negative  ions.  The 
position  of  oxygen  in  this  respect  is  harder  to  define,  for  no  simple 
ions  of  oxygen  exist  in  solution.  Under  certain  conditions  chlo- 
rine displaces  oxygen  from  water,  as  in  Exp.  10,  Chap.  II;  yet 
some  substances  which  give  up  oxygen  readily  (oxidizing  agents, 
see  Exp.  10)  set  chlorine  free  from  hydrochloric  acid. 

17.  Observe  the  color  of  the  solution  of  hydriodic  acid 
made  in  Preparation  8,  or  of  the  stock  solution  on  the  side 
shelf.     If  this  should  chance  to  be  perfectly  fresh  and  color- 
less, pour  a  few  cubic  centimeters  into  the  bottom  of  a  beaker 
and  let  it  stand  some  time  exposed  to  the  air.     Apply  tests 
to  determine  if  the  color  is  caused  by  free  iodine. 

Are  hydrobromic  and  hydrochloric  acids  similarly  affected 
by  exposure  to  the  air?  Is  a  solution  of  potassium  iodide  so 
affected? 

Hydriodic  acid  solution  can  be  prepared  perfectly  colorless  and 
kept  colorless  so  long  as  it  is  kept  out  of  contact  with  the  air. 
In  contact  with  air  it  becomes  brown  in  a  very  few  minutes  and 
the  brown  color  gradually  increases  in  depth.  This  color  is  due 
to  free  iodine  which  is  quite  soluble  in  an  iodide  solution  giving  a 
brown  solution 

2H+  +  21-  +  i02  -»  H20  +  I2. 

In  the  course  of  this  reaction  hydrogen  ions  are  used  up;  it  is 
therefore  obvious  that  the  presence  of  hydrogen  ions  will  aid  in 
the  displacement  of  iodine  ions  by  oxygen.  In  a  neutral  potas- 
sium iodide  solution  no  iodine  is  observed  and  it  is  thus  apparent 
that  the  help  of  hydrogen  ions  is  necessary  to  accomplish  the 
liberation  of  iodine. 

Chlorine  and  bromine  are  not  liberated  by  the  action  of  air  on 
hydrochloric  acid  and  hydrobromic  acid,  still  less  on  neutral 
solutions  of  chlorides  and  bromides. 

18.  Add  a  few  drops  of  iodine  solution  to  a  little  hydrogen 
sulphide  water.     The  brown  iodine  solution  is  immediately 
decolorized  and  a  white  cloudiness  (precipitate  of  sulphur) 
appears  in  the  solution. 


ACTIVITY  OF  THE  HALOGENS,  ETC. 


151 


Observe  the  precipitate  which  slowly  forms  in  the  bottle 
of  hydrogen  sulphide  water  to  which  air  has  some  access. 

Explain  the  reaction  in  each  case,  and  place  sulphur, 
oxygen,  chlorine,  bromine,  and  iodine  in  the  order  of  their 
chemical  activity  in  acid  solutions. 

From  these  experiments  it  is  seen  that  both  iodine  and  oxygen 
are  more  active  than  sulphur.  The  reactions  are  quite  certainly 
ionic  displacements  although  it  is  rather  difficult  to  represent  them 
as  such. 

H2S    -*     2H+    +    S- 
I2 

t- 

SI 


H2S 


H2O 


2H  + 

+  s- 

Y 

o- 

si 


Hydrogen  sulphide,  although  its  ionization  is  very  small,  does 
maintain  a  certain  S~"~  ion  concentration,  and,  as  rapidly  as 
the  S  —  ion  is  driven  out,  the  hydrogen  sulphide  ionizes  further. 
Thus  all  of  the  sulphide  constituent  of  the  hydrogen  sulphide 
enters  the  reaction  and  is  precipitated  as  free  sulphur.  Suppos- 
edly oxygen  strives  to  form  the  hypothetical  O  ions  but  this 
unites  with  hydrogen  ions  to  form  water. 

When  one  is  sufficiently  familiar  with  the  ionic  theory  he  writes 
the  equations  for  these  reactions  in  the  abbreviated  form 

H2S  +  I2^2H+      21-  +S 

H2S  +  JO2  ->  H20  +  S 


but  he  can  read  in  these  equations  the  whole  of  the  above  dis- 
cussion. 

From  this  experiment  and  the  preceding  one  we  have  learned 
that  the  non-metallic  elements  fall  in  the  order  F,  Cl,  Br,  O,  I,  S, 
(fluorine  being  strongest)  with  respect  to  their  activity  in  acid 
aqueous  solution.  This  is  the  order  of  the  electromotive  series. 


152      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

If  the  solution  is  made  neutral  the  electromotive  potential  of 
oxygen  is  lowered  so  that  the  oxygen  is  no  longer  able  to  displace 
iodine. 

In  dry  gaseous  mixtures  the  order  of  activity  of  the  non-metals 
is  somewhat  altered,  oxygens  moving  up  ahead  of  bromine  and 
chlorine,  and  sulphur  and  iddine  changing  places:  F,  O,  Cl,  Br, 
8,1. 

SULPHUR 

19.  Allotropic  Forms,     (a)  Dissolve  in  a  dry  test  tube  a 
small  piece  of  roll  sulphur  in  3  to  5  cc.  of  carbon  disulphide. 
Pour  the  clear  solution  on  a  watch  glass  and  allow  it  to  evap- 
orate spontaneously  under  the  hood.     Examine  the  crystals. 

(6)  Melt  50  grams  of  sulphur  in  a  small  beaker.  Get  a 
special  beaker  from  the  Supply  Room.  After  the  experi- 
ment throw  it  away.  It  cannot  be  cleaned.  Then  allow 
the  melt  to  cool  very  slowly,  and  when  it  has  partially  solidi- 
fied, and  a  crust  has  formed  over  the  surface,  break  this 
crust  and  pour  out  what  is  still  left  in  the  liquid  state.  Ex- 
amine the  crystals  and  compare  them  with  those  observed 
in  (a). 

(c)  Melt  10  grams  of  sulphur  in  a  test  tube.  Heat  slowly, 
and  observe  all  the  changes  that  take  place  during  the  heating. 
When  the  sulphur  boils,  pour  it  into  a  pan  or  large  beaker 
of  water  and  observe  the  condition  of  the  cooled  product. 

Distinguish  between  the  three  solid  forms  of  sulphur  here 
observed  and  discuss  the  differences  in  the  conditions  under 
which  they  are  formed  (see  reference  book). 

20.  Hydrogen  and   Sulphur,     (a)  Prepare  hydrogen  sul- 
pliide  by  the  action  of  dilute  sulphuric  acid  on  ferrous  sul- 
phide in  a  generator  bottle  (see  Note  13(a)  page  19).     Pass 
the  gas  through  a  drying  tube  containing  a  plug  of  cotton 
wool,  which  retains  any  acid  spray,  then  through  a  piece  of 
hard  glass  tube  the  end  of  which  is  drawn  out  into  a  capillary 
jet.     Heat  a  section  of  the  tube  to  redness  and  observe  that 
a  white  deposit  is  formed  on  the  walls  of  the  cooler  part  of 
the  tube  beyond. 

(6)  Light  the  gas  issuing  from  the  capillary  jet  and  note 
the  products  of  the  freely  burning  hydrogen  sulphide.  The 
odor  of  sulphur  dioxide  is  unmistakable.  If  a  cold  bottle 


SULPHUR  153 

is  held  over  the  flame,  dew  condenses  inside  the  bottle.  Now 
thrust  a  piece  of  cold  procelain  half  way  into  the  flame  and 
note  the  deposit  of  sulphur. 

(c)  Generate  hydrogen  gas  in  exactly  the  same  way  as 
hydrogen  sulphide  in  (a).  Place  about  \  gram  of  sulphur 
in  the  end  of  the  hard  glass  tube  nearest  the  generator. 
When  it  is  ascertained  that  the  hydrogen  coming  off  is  pure, 
heat  the  further  end  of  the  glass  tube  to  redness  and  gradually 
move  the  flame  towards  the  sulphur  until  this  begins  to  be 
volatilized  a  little.  By  this  arrangement  a  mixture  of  hydro- 
gen and  sulphur  vapor  is  made  to  pass  through  a  red-hot 
tube.  Test  the  escaping  gases  with  lead  acetate  paper. 

Parts  (a)  and  (6)  show  that  sulphur  is  deposited  when  hydrogen 
sulphide  is  strongly  heated.  In  the  freely  burning  flame  the 
ultimate  products  are  sulphur  dioxide  and  water  : 

H2S  +  f  02  ->  H20  +  S02. 

But  if  the  unburned  vapors  in  the  interior  of  the  flame  are  cooled 
by  the  porcelain  before  they  can  burn  we  find  that  sulphur  is 
present  there.  Although  we  have  not  directly  proved  the  presence 
of  hydrogen  we  are  pretty  well  justified  in  thinking  that  it  is  the 
other  decomposition  product  of  hydrogen  sulphide. 

Part  (c)  shows  that  some  hydrogen  sulphide  is  formed  when  a 
mixture  of  hydrogen  and  sulphur  vapor  is  heated  to  redness  and 
it  is  thus  clear  that  neither  the  synthesis  nor  the  decomposition 
of  hydrogen  sulphide  is  complete  at  this  temperature  but  that 
hydrogen  sulphide  reaches  an  equilibrium  with  its  products  ac- 
cording to  the  reversible  reaction. 


It  is  instructive  to  review  the  properties  of  the  other  non- 
metals  by  considering  how  they  would  behave  in  similar  circum- 
stances. 

Chlorine  and  hydrogen,  passed  together  into  a  heated  tube,  com- 
bine completely  and  with  explosive  violence;  oxygen  and  hydro- 
gen explode  even  more  violently;  and  fluorine  and  hydrogen  can 
hardly  be  mixed  together,  even  at  ordinary  temperatures  with- 
out exploding.  Bromine  and  hydrogen  led  through  a  heated  tube 
combine  freely  to  form  hydrogen  bromide  but  without  any  ex- 


154     NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

plosion.     Iodine  and  hydrogen  combine  to  but  a  limited  extent, 
-  less  than  sulphur  and  hydrogen. 

On  the  other  hand  the  chemical  activity  of  the  non-metal  may 
be  judged  by  the  stability  of  the  hydrogen  compound,  and  the 
stability  in  turn  may  be  measured  by  the  temperature  to  which 
it  must  be  heated  before  it  is  perceptibly  dissociated.  Hydrogen 
iodine  begins  to  show  the  violet  color  of  free  iodine  as  low  as 
180°;  hydrogen  sulphide  does  not  deposit  fre,e  sulphur  lower  than 
310°;  hydrogen  bromide  is  first  perceptibly  decomposed  at  800°; 
hydrogen  chloride  at  1800°;  water  at  2000°;  and  hydrogen 
fluoride  is  still  undissociated  at  this  temperature. 

21.  lonization  of  Hydrogen  Sulphide.     Pass  hydrogen  sul- 
phide, which  has  been  strained  through  cotton  wool,  and 
which  has   bubbled   through   one  bottle   containing  water, 
into  distilled  water  in  a  flask  until  the  solution  is  saturated. 
Test  the  conductivity  of  the  solution  with  Electrodes  C  (see 
page  70)  and  test  it  with  litmus.     The  conductivity  is  almost 
imperceptible;   it  does  not  cause  the  lamp  to  glow,  but  it  is 
shown  by  a  slight  evolution  of  bubbles  from  the  electrodes. 
Litmus  is  turned  towards  the  red,  but  not  the  full  red  color 
produced  by  strong  acids. 

The  effects  observed  in  this  experiment  are  caused  by  the 
ionization  H2S;=±H+  +  HS~  to  the  extent  of  0.05  per  cent  as 
found  in  the  tables. 

22.  Hydrogen  Sulphide  as  a  Precipitant.     Pass  hydrogen 
sulphide  into  solutions,  acidified  or  neutral,  of  salts  of  the 
heavy  metals;  take  for  example  a  solution  of  copper  chloride. 

A  heavy  black  precipitate  is  formed. 

The  precipitate  is  copper  sulphide.  The  table  gives  the  ioniz- 
ation HS-^±H+  +  S —  as  0.0002  per  cent  but  this  applies  only 
to  a  0.1  equivalent  solution  of  Na+  HS~  in  which  all  the  H+  ions 
come  from  the  dissociation  in  question.  In  a  solution  of  H2S  con- 
taining the  H+  ions  of  the  first  dissociation,  the  concentration  of 
S —  ions  would  be  very  much  smaller.  In  a  solution  containing  a 
strong  acid  like  HC1,  the  concentration  of  S —  which  could  come 
from  H2S  would  be  excessively  small.  But  copper  sulphide  is 
precipitated  from  such  a  solution.  The  solubility  product  of 
copper  sulphide  must,  then  be  extremely  small. 


NITROGEN  155 

The  solubility  of  all  the  heavy  metal  sulphides  is  very  small; 
they  are  all  insoluble  in  water,  but  some  dissolve  in  hydrochloric 
acid.  The  solubility  product  of  these  is  not  quite  so  small  because 
it  is  not  reached  when  the  ionization  of  H2S  is  driven  back  by  the 
strong  acid. 

23.  Reducing  Action  of  Hydrogen  Sulphide.     Dilute  5  cc. 
of  a  potassium  permanganate  solution  with  100  cc.  of  water 
and  add  5  cc.  of  dilute  sulphuric  acid.     Bring  the  solution 
to  boiling  and  pass  in  hydrogen  sulphide.     The  deep  purple 
color  of  the  permanganate  quickly  disappears,  and  a  white 
opalescent  precipitate  appears. 

Although  sulphur  in  hydrogen  sulphide  has  a  range  of  eight  pos- 
sible points  in  valence, —  from  —  II  to  +  VI, —  to  go  when  it  acts 
as  a  reducing  agent,  it  usually  does  not  go  beyond  the  zero  stage 
of  valence  in  aqueous  oxidizing  agents,  the  obvious  explanation 
being  that  the  free  sulphur  is  precipitated  and  thus  removed  from 
the  sphere  of  action. 

2KMnO4  +  5H2S  +  3H2S04  7*  K2S04  +  2MnS04  +  8H20  +  5S 
2Mn  +  VII  to  +  II          2  X  (-  5)  =  -  10 

5S  -  II     to       0  5  X  (+  2)  =  +  10 

Total  change  =     0 

NITROGEN 

Nitrogen  is  an  extremely  inactive  element,  combining  directly 
only  with  the  most  active  of  the  metals.  Nitrogen  forms  no 
simple  ions  and  it  is  impossible  to  give  it  an  exact  potential  in  the 
electromotive  series,  but  it  is  obvious  enough  that  it  is  far  less 
active  than  sulphur  or  iodine.  Under  the  influence  of  electric 
sparks  or  of  catalyzers  at  about  400°  nitrogen  does  combine 
sparingly  with  hydrogen  to  form  ammonia  NH3.  It  combines 
quite  readily  with  magnesium  to  give  the  nitride  Mg3N2  which 
compound  bears  the  same  relation  to  ammonia  (its  formula  should 
logically  be  H3N)  that  magnesium  chloride  does  to  hydrogen 
chloride. 

24.  Synthesis  of  Ammonia.     Heat  a  mixture  of  14  grams 
of  iron  filings,  0.5  gram  of  powdered  sodium  hydroxide,  and 
0.5  gram  of  powdered  potassium  nitrate  in  a  test  tube,  and 
test  the  gas  which  escapes  by  holding  a  rod  wet  with  con- 


156      NON-METALLIC  ELEMENTS  IN  BINARY  COMPOUNDS 

centrated  hydrochloric  acid  near  the  mouth  of  the  tube. 
A  white  smoke  is  formed. 

Test  the  action  of  iron  on  sodium  hydroxide  and  potassium 
nitrate  respectively  by  heating  7  grams  portions  of  iron 
filings  with  0.5  gram  ofUeach  of  these  reagents  separately. 
From  sodium  hydroxide,  a  gas  that  burns  with  a  colorless 
flame  (hydrogen)  is  evolved.  From  potassium  nitrate  a 
gas  that  will  neither  burn  nor  support  combustion  (nitrogen) 
is  evolved. 

Iron  is  more  active  than  either  hydrogen  or  nitrogen  and  dis- 
places them 

2KN03  +  5Fe  -»  K20.5FeO  +  2N 
2NaOH  +  Fe  -»  Na2O.FeO  +  2H 

The  hydrogen  and  nitrogen  are  doubtless  in  the  atomic  state  at 
the  moment  of  liberation.  They  change  at  once  to  ordinary 
molecular  N2  and  H2  if  they  find  nothing  to  combine  with,  but 
in  the  presence  of  each  other  they  combine  to  form  ammonia. 

25.  Nitrides.  Place  half  a  gram  of  magnesium  ribbon 
rolled  into  a  ball,  or  half  a  gram  of  powdered  magnesium  in 
a  small  crucible.  Heat  it  until  it  catches  fire,  put  the  cover 
.on  to  restrict  the  admission  of  air  and  let  the  magnesium 
slowly  burn.  Remove  the  ash  to  a  test  tube  and  add  a  few 
drops  of  water  (cautiously).  Note  the  odor  of  ammonia, 
and  bring  a  rod  wet  with  concentrated  hydrochloric  acid  near 
the  mouth  of  the  tube,  noting  the  white  smoke. 

When  enough  oxygen  cannot  come  into  contact  with  burning 
magnesium  to  form  the  oxide,  magnesium  combines  readily  with 
nitrogen.  Of  course  prolonged  heating  in  air  will  convert  all 
nitride  into  oxide,  but  if  the  ash  is  cooled  at  once  it  contains 
a  considerable  amount  of  nitride.  Magnesium  nitride  hydrolyzes 
very  easily  according  to  the  reaction 

Mg3N2  +  6H2O  -*  3Mg(OH)2  +  2H3N. 

The  magnesium  nitride  may  be  considered  as  a  salt  of  the  base 
Mg(OH)2  and  the  acid  H3N.  But  since  H3N  is  so  weak  an  acid 
that  it  is  not  usually  considered  as  an  acid  at  all,  the  hydrolysis  of 
its  salt  is  correspondingly  very  complete. 


GENERAL  QUESTIONS  157 

GENERAL  QUESTIONS  IV 

1.  What  are  the  distinctive  physical  and  the  distinctive  chemi- 
cal properties  of  the  non-metallic  elements? 

2.  Why  are  the  non-metallic  elements  considered  to  be   the 
negative  constituents  of  binary  compounds? 

3.  Many  binary   compounds   of   non-metals  with  non-metals 
are  known,  but  such  compounds  usually  can  exist  only  out  of 
contact    with    water.     For    example,    phosphorus    trichloride    is 
completely  hydrolyzed  by  water.     Write  the  equation  for  this 
reaction,  and  treating  it  as  a  metathesis,  conclude  which  element 
in  the  phosphorus  trichloride  is  to  be  regarded  as  the  positive 
constituent. 

4.  What  is  meant  by  the  "  activity  "  of  an  element.     Arrange 
the  non-metals  studied  in  this  chapter  in  the  order  of  their  activity 
as  negative  elements,  (a)  when  they  react  in  aqueous  solution, 
(6)  when  they  react  with  "  dry  "  substances. 

5.  Look  up  the  heat  of  solution  of  hydrogen  iodide  and  hydrogen 
sulphide,  and  state  how  this  factor  is  able  to  make  iodine  more 
active  than  sulphur  in  aqueous  solution,  whereas  sulphur  is  the 
more  active  in  the  dry  state. 


CHAPTER  V 
ALKALI  AND   ALKALINE  EARTH  METALS 

These  metals  constitute  the  left  hand  or  A  families  in  Groups  I 
and  II  of  the  periodic  classification  of  the  elements,  as  shown  in 
the  table  inside  the  front  cover  of  the  book. 

The  metals  of  these  two  families  are  studied  together  because 
they  are  extremely  active  base-forming  elements.  On  account 
of  their  great  activity  they  are  never  found  uncombined  in  nature, 
and  it  is  only  by  the  aid  of  the  most  powerful  reducing  agencies 
(for  example,  by  electrolysis  of  their  molten  salts)  that  the  metals 
themselves  are  extracted  from  their  compounds. 

The  alkali  metals  are  monovalent.  Their  hydroxides,  MOH, 
are  extremely  soluble  and  are  highly  ionized  as  bases;  on  account 
of  the  corrosive  properties  of  the  latter  they  are  known  as  the 
caustic  alkalies  —  hence  the  designation,  alkali  metals.  The 
compounds  of  the  alkali  metals  are,  with  a  very  few  exceptions, 
soluble  in  water,  and  they  are  all  strong  electrolytes. 

The  radical  ammonium,  NH4,  is  classed  with  the  alkali  metals 
on  account  of  its  ability  to  form  the  same  kinds  of  compounds. 

The  alkaline  earth  metals  are  divalent;  their  hydroxides, 
M(OH)2,  are  less  soluble  than  those  of  the  alkali  metals,  but  are 
nevertheless  very  strongly  basic.  The  compounds  of  these 
metals  are  not  so  generally  soluble  as  those  of  the  alkali  metals, 
and  in  particular  the  carbonates,  sulphates  and  phosphates  are 
mostly  insoluble. 

PREPARATION  15 
SODIUM  BICARBONATE  BY  THE  AMMONIA  (SOLVAY)  PROCESS 

Materials:     table  salt,  59  grams; 

14-normal  ammonium  hydroxide,  71  cc. 
carbon  dioxide,  which  can  be  drawn  either  from  an 
automatic  gas  generator  (see  Note  13 (c)  on  page 
20),  or  from  a  gas  holder  which  is  filled  as  needed 
from  a  steel  cylinder  of  liquid  carbon  dioxide. 
158 


SODIUM   BICARBONATE  159 

Apparatus:  bubbling  bottle,  through  which  the  carbon  dioxide 
is  to  pass  to  indicate  its  rate  of  flow. 

suction  filter  (see  Note  4,  (a)  and  (b),  p.  5). 

1-liter  flask  equipped  with  1-hole  rubber  stopper, 
delivery  tube  reaching  to  bottom,  and  24  inches 
of  rubber  delivery  tube. 

300-cc.  flask  with  stopper. 

Procedure:  Place  the  salt,  the  ammonium  hydroxide,  and  130  cc. 
of  water  in  the  smaller  flask  and  shake  vigorously  until  the  salt 
is  dissolved.  Pour  the  solution  through  a  filter  into  the  1-liter 
flask  (large  plaited  filter  for  speed).  Use  this  flask  as  the  absorp- 
tion vessel  and  connect  it  with  the  source  of  carbon  dioxide. 
Loosen  the  stopper  of  the  flask  and  let  the  carbon  dioxide  expel 
the  air.  Then  stopper  the  flask  and  shake  it  as  vigorously  as 
possible  until  the  absorption  of  carbon  dioxide  has  slackened. 
Loosen  the  stopper  again  to  allow  any  air  that  has  accumulated 
in  the  flask  to  escape;  then  continue  the  shaking  until  practically 
no  more  gas  passes  the  bubbling  bottle  even  with  vigorous  shaking. 
If  the  shaking  has  been  continuous,  this  point  will  be  reached 
within  thirty  minutes.  When  the  absorption  is  complete  collect 
the  precipitated  sodium  bicarbonate  on  the  suction  filter  (Notes  3, 
and  4,  b),  drain  it  thoroughly  with  suction,  stop  the  suction,  pour 
over  the  surface  of  the  product  15  cc.  of  cold  water,  and  after  this 
has  soaked  in  apply  the  suction  again.  Wash  a  second  time  with 
15  cc.  of  cold  water  exactly  as  at  first.  Spread  the  drained  product 
on  paper  towels  and  leave  it  at  room  temperature  24  hours  to  dry. 
Test  the  preparation  for  chloride  by  dissolving  about  0.1  gram  in 
a  little  water,  acidulating  slightly  with  nitric  acid,  and  adding  a 
drop  of  silver  nitrate  solution.  There  will  be  considerable  clouding. 
Put  up  the  product  in  a  2-ounce  cork  stoppered  bottle. 

QUESTIONS 

1.  What  is  the  purpose  of  washing  the  product  with  water? 
How  much  sodium  bicarbonate  is  lost  in  this  way  (see  solubility 
table)? 

2.  Why  must  the  solution  be  acidulated  with  nitric  acid  before 
testing  with  silver  nitrate? 

3.  Why  does  shaking  greatly  increase  the  rate  of  absorption? 


160  ALKALI  AND  ALKALINE  EARTH   METALS 

4.  How  do  you  explain  the  heat  produced  in  the  absorption 
flask? 

5.  How  can  you  prepare  sodium  carbonate  from  sodium  bi- 
carbonate? 

6.  Why  cannot  potassium  bicarbonate  be  effectively  prepared 
from  potassium  chloride  by  the  ammonia   process?     (Look   up 
the   solubility  of   potassium  bicarbonate.)     What   process  may 
be  used  to  prepare  potassium  carbonate  from  this  source? 

7.  What  is  an  acid  salt?     How  does  a  solution  of  an  acid  salt 
such  as  KHS04  behave  toward  litmus?     Test  the  behavior  of 
solutions    of    NaHC03    and    Na2CO3    towards    litmus.     Explain 
the  cause  of  this  behavior. 

8.  Would  a  precipitate  of  sodium  bicarbonate  form  if  carbon 
dioxide  were  passed  into  a  solution  of  sodium  chloride  alone? 
Explain  the  part  played  by  the  ammonia  in  the  formation  of  the 
product,  giving  the  ionic  equation. 

9.  Explain  how  a  given  amount  of  ammonia  may  be  used  over 
and  over  again  in  the  commercial  process. 

PREPARATION  16 
SODIUM  CARBONATE  FROM  SODIUM  BICARBONATE 

Heat  the  sodium  bicarbonate  obtained  in  No.  15  in  a  4-inch 
evaporating  dish  by  playing  the  free  flame  over  the  bottom  of  the 
dish  until  carbon  dioxide  ceases  to  escape.  Do  not  set  the  burner 
down  under  the  dish  because  the  flame  concentrated  in  one  spot 
might  melt  the  sodium  carbonate.  Compare  the  weight  obtained 
with  that  calculated.  Put  up  the  product  in  a  6-inch  cork-stop- 
pered test  tube. 

PREPARATION  17 
CAUSTIC  ALKALI  FROM  ALKALI  CARBONATE 

Of  the  mineral  constituents  of  plants,  potassium  salts  form 
an  important  part,  and,  so  far  as  "these  are  salts  of  organic  acids, 
they  are  converted  into  potassium  carbonate  when  the  plant  is 
burned.  On  an  average,  wood  ashes  contain  about  10  percent 
of  potassium  carbonate,  and  before  the  advent  of  the  Leblanc 
soda  process  this  was  almost  the  sole  supply  of  alkali.  Even  after 
this  process  came  into  general  use,  by  which  sodium  carbonate 


CAUSTIC  ALKALI  FROM  ALKALI  CARBONATE    161 

could  be  obtained  from  common  salt,  wood  ashes  remained  for 
some  time  the  important  source  of  potassium  carbonate.  In 
recent  years,  however,  the  greater  part  of  the  production  of 
potassium  carbonate  has  been  from  potassium  salts  found  in 
deposits  in  the  earth,  principally  at  Stassfurt,  Germany. 

Potassium  carbonate  being  the  principle  soluble  constituent 
of  wood  ashes,  it  is  extracted  with  water;  but  the  extract  so 
obtained  contains,  as  well,  the  other  soluble  mineral  constituents, 
and  also  a  considerable  amount  of  tarry  coloring  matter  which 
was  not  destroyed  in  the  combustion  of  the  wood.  This  tarry 
matter  is  destroyed  by  calcination  of  the  residue  obtained  on 
evaporating  the  aqueous  extract,  and  the  calcined  mass  is  what  is 
known  as  crude  potash.  A  better  grade  of  commercial  potash 
can  be  obtained  by  dissolving  this  mass  in  water,  filtering,  and 
evaporating  the  solution. 

In  order  to  obtain  the  hydroxide  or  caustic  alkali  from  potassium 
carbonate  or  sodium  carbonate,  the  carbonate  in  10  percent 
solution  is  treated  with  excess  of  milk  of  lime  (calcium  hydroxide), 
the  filtrate  being  a  pure  solution  of  the  desired  hydroxide.  This 
process  involves  a  most  important  application  of  the  principle  of 
solubility  product:  calcium  hydroxide  continues  to  dissolve  and 
calcium  carbonate  to  precipitate  according  to  the  reaction 

Ca(OH)2    ^±    Ca++    +    20H~ 
C08—  +    2K+ 

\\ 
CaC03  | 

until  equilibrium  between  both  solids  and  the  solution  is  attained. 


Solubility  in  pure  water 
moles  per  liter 

Solubility  product 

Ca(OH)2  .  , 
CaCO3  .... 

0.02 

0.00013 

.02X  .042 
.  00013  X  .00013 

At  this  point  the  concentration  of  both  Ca++  and  C03  ions 
is  so  small  (both  close  to  the  value  .00013)  that  for  practical  pur- 
poses it  is  disregarded  and  the  filtered  solution  is  said  to  be  free 
of  either  calcium  or  carbonate.  But  with  so  small  a  Ca++  ion 
concentration  the  OH~  ion  concentration  can  be  quite  large 


162  ALKALI  AND  ALKALINE  EARTH  METALS 

without  exceeding  the  solubility  product  of  Ca(OH)2.  In  fact 
the  filtered  solution  has  a  concentration  of  nearly  2-molal  in 
KOH.  It  would  obviously  be  more  economical  to  start  with  a 
much  more  concentrated  K2CO3  solution,  since  a  more  concen- 
trated KOH  solution  would  jequire  less  evaporating  to  give  the 
solid  potassium  hydroxide.  When,  however,  this  is  done  the  high 
OH~  ion  concentration  lessens  the  equilibrium  Ca++  ion  concen- 
tration which  in  turn  increases  the  equilibrium  C03  ion  con- 
centration. In  other  words  if  a  solution  of  alkali  carbonate  more 
concentrated  than  10  percent  is  taken  at  the  start,  the  conversion 
into  alkali  hydroxide  will  be  incomplete. 

Materials:     anhydrous  sodium  carbonate,  Na^COs,  53  grams  = 
i  F.W.  (or  any  alkali  carbonate  as  for  example  the 
aqueous  extract  from  wood  ashes.) 
slaked  lime,  Ca(OH)2,  about  50  grams. 
Apparatus:  8-inch  dish, 
suction  filter, 
burette  with  normal  HC1. 
15-cc.  pipette. 
500-cc.  bottle  with  rubber  stopper. 

Procedure:  Dissolve  the  sodium  carbonate  in  300  cc.  of  water, 
and  stir  the  slaked  lime  into  another  300  cc.  of  water,  making  milk 
of  lime.  Bring  the  carbonate  solution  to  boiling  in  the  8-inch 
dish  and  pour  the  milk  of  lime  slowly  with  stirring  into  the  boiling 
solution.  Let  the  mixture  boil  15  minutes  and  then  filter,  using 
a  suction  bottle  (see  Note  4  (6)).  Measure  the  volume  of  the  solu- 
tion of  caustic  alkali  obtained  and  preserve  it  in  the  rubber-stop- 
pered bottle. 

Test  the  strength  of  the  solution.  Measure  15  cc.  with  a  pipette 
into  a  beaker  and  add  a  drop  of  litmus  solution.  Run  into  this 
from  a  burette  a  solution  of  normal  hydrochloric  acid,  drop  by 
drop,  until  the  color  just  changes  from  blue  to  red.  If  the  right 
point  is  overstepped  begin  again  with  a  fresh  sample  of  the  solution. 
From  the  amount  of  acid  taken  to  neutralize  the  sample,  calculate 
the  amount  of  alkali  metal  hydroxide.  Label  the  bottle  with 
the  number  of  cubic  centimeters  of  the  solution,  with  its  strength 
in  mols  per  liter,  and  with  the  actual  amount  in  grams  of  the 
hydroxide. 


CHEMICALLY  PURE  SODIUM   CHLORIDE  163 

QUESTIONS 

1.  The  calcium  hydroxide  used  in  causticizing  is  soluble  only 
to  the  extent  of  1.7  grams  per  liter.     Explain  how,  in  spite  of  its 
limited  solubility,  the  required  amount  can  enter  into  reaction. 

2.  Explain  why  the  solution  obtained  contains  practically  no 
calcium  although  an  excess  of  calcium  hydroxide,  which  is  appre- 
ciably soluble  in  pure  water,  has  been  used  for  causticizing. 

PREPARATION  18 

CHEMICALLY  PURE  SODIUM  CHLORIDE  FROM  ROCK  SALT 
Common  rock  salt  may  contain  other  than  sodium  chloride 
up  to  10  percent  of  matter,  which  consists  in  the  main  of  the 
sulphates  and  chlorides  of  potassium,  calcium,  and  magnesium, 
not  to  mention  a  considerable  amount  of  dirt  and  insoluble  matter. 
For  most  commercial  purposes  these  impurities  are  not  harmful. 
By  careful  crystallization  of  the  salt  from  solution,  a  product 
sufficiently  free  from  these  impurities  can  be  obtained  to  be  used 
as  table  salt.  To  obtain  chemically  pure  sodium  chloride,  however, 
more  elaborate  precautions  must  be  taken.  A  satisfactory  method 
depends  upon  the  insolubility  of  sodium  chloride  in  a  concentrated 
solution  of  hydrochloric  acid.  A  nearly  saturated  solution  of  the 
rock  salt  is  prepared,  and,  without  removing  the  dirt  and  insoluble 
matter,  enough  pure  sodium  carbonate  is  added  to  precipitate  the 
calcium  and  magnesium  in  the  solution  as  carbonates.  Into  the 
clear  filtrate  is  then  passed  gaseous  hydrochloric  acid  until  the 
greater  part  of  the  sodium  chloride  is  precipitated,  while  the  small 
amounts  of  sulphates  and  of  potassium  salts  remain  in  the  solution. 
The  precipitate  is  drained  and  washed  with  a  solution  of  hydro- 
chloric acid  until  the  liquid  clinging  to  the  crystals  is  entirely  free 
from  sulphates. 

Materials:     rock  salt,  75  grams. 

concentrated  sulphuric  acid,  95  cc. 

concentrated  hydrochloric  acid,  50  cc. 
Apparatus:  1  liter  r.  b.  flask  with  2-hole  rubber  stopper. 

300  cc.  common  bottle  with  3-hole  rubber  stopper. 

2-inch  funnel. 

2  thistle  tubes. 

delivery  tube  and  connection,  see  diagram. 

beaker,  3  inches  wide. 


164 


ALKALI  AND  ALKALINE  EARTH  METALS 


Procedure:  Dissolve  25  grams  of  rock  salt  in  75  cc.  of  water, 
hastening  the  action  with  gentle  heating.  To  the  solution  add 
about  1  gram  of  sodium  carbonate  dissolved  in  a  few  cubic  centi- 
meters of  water.  Stir,  let  settle,  and  add  a  few  drops  more  of 
sodium  carbonate  solution,  and  if  no  fresh  precipitate  is  pro- 


FIG.  19 

duced  in  the  clear  part  of  the  solution  no  more  need  be  added; 
otherwise  enough  more  must  be  added  to  produce  this  result. 
Filter  the  solution,  hot,  through  an  ordinary  filter  (Note  4  (c)). 
Prepare  pure  hydrogen  chloride  by  placing  50  grams  of  rock  salt 
in  the  generator  flask  and  treating  it  with  the  95  cc.  of  concentrated 
sulphuric  acid.  Observe  the  direction  given  under  the  preparations 
of  hydrochloric  acid,  page  119,  and  note  particularly  the  caution  as 
to  the  disposal  of  the  hot  residue.  The  gas  is  to  be  purified  by 
being  bubbled  through  the  washing  bottle  containing  about  50  cc. 
of  concentrated  hydrochloric  acid,  and  then  it  is  passed  through 


AMMONIUM   BROMIDE  165 

the  wide  mouth  funnel  into  the  sodium  chloride  solution  in  the 
3-inch  beaker.  When  hydrogen  chloride  ceases  to  be  evolved  on 
heating  the  generator,  separate  the  precipitated  sodium  chloride 
in  a  suction  filter  from  its  mother  liquor.  Test  the  filtrate  for 
sulphate  by  adding  a  little  barium  chloride  solution  to  a  small 
sample  of  it  diluted  with  water.  A  strong  test  will  probably  be 
obtained.  Now  wash  the  crystals  with  successive  portions  of 
10  cc.  of  6-n  HC1,  until  the  washings  show  no  further  test  for 
sulphates.  (See  Note  5  (a).)  Then  transfer  the  crystals  to  a 
porcelain  dish  and  heat  gently,  while  stirring,  until  all  decrepita- 
tion ceases.  Put  up  the  product  in  a  6-inch  cork-stoppered  test 
tube. 

QUESTIONS 

1.  Why  must  the  hydrochloric  acid  gas  be  passed  through  a 
washing  bottle?     Why  is  the  safety  tube  necessary? 

2.  Why,  in  the  light  of  the  law  of  molecular  concentration, 
should  one  expect  the  solubility  of  sodium  chloride  to  be  lessened 
by  the  presence  of  hydrochloric  acid?     It  may  be  stated  that 
another  effect  known  as  the  "  salting  out  effect  "  also  comes  into 
play  here  and  likewise  tends  to  lessen  the  solubility  of  sodium 
chloride.     The  great  amount  of  heat  liberated  when  hydrogen 
chloride  dissolves  in  water  indicates  a  chemical  action,  and  it  is 
very  probable  that  the  water  and  hydrogen  chloride  unite  to  form 
an  unstable  compound.     In  the  saturated  solution  then  nearly  all 
the  water  is  chemically  combined  and  very  little  is  left  to  hold 
sodium  chloride  in  solution. 

3.  Mention   two   possible   causes   for   the   very   considerable 
amount  of  heat  produced  when  the  hydrochloric  acid  gas  is  ab- 
sorbed by  the  solution  in  the  beaker. 

4.  Why  does  not  the  solution  in  the  washing  bottle  also  grow 
hot? 

PREPARATION  19 
AMMONIUM  BROMIDE 

Ammonium  bromide  could  be  prepared  by  the  neutralization 
of  ammonium  hydroxide  with  hydrobromic  acid, 
NH4OH  +  HBr  =  NH4Br  +  H2O. 

Since,  however,  hydrobromic  acid  is  a  more  expensive  material 
than  uncombined  bromine,  the  latter  would  have  the  preference 


166  ALKALI  AND  ALKALINE  EARTH   METALS 

as  a  source  of  bromine,  provided  it  yielded  as  satisfactory  a  prod- 
uct. Chlorine  or  bromine  reacts  as  follows  upon  a  cold  solution 
of  sodium  hydroxide,  as,  for  example,  in  the  manufacture  of 
bleaching  liquors, 

Br2  +  2NaOH  ^NaBr  +  NaBrO  +  H20, 


with  the  formation  of  sodium  hypochlorite  or  hypobromite. 
Sodium  hypobromite  reacts  with  ammonium  hydroxide  according 
to  the  equation, 

3NaBrO  +  2NH4OH  =  3NaBi  +  5H2O  +  N2. 

If  now  we  substitute  NH4  for  Na  in  the  last  two  equations  and 
add  the  equations  we  obtain 

3Br2  +  8NH4OH  ->  6NH4Br  +  N2  +  8H2O 

which  gives  the  complete  reaction  of  bromine  with  ammonium 
hydroxide. 

Materials:     bromine,  Br2,  40  grams  =  12.5  cc.  =  J  F.W. 

NH4OH  15-normal  =  44  cc.  =  f  F.W. 
Apparatus:  50  cc.  dropping  funnel. 

300  cc.  Erlenmeyer  flask. 

4-inch  evaporating  dish. 

pan  of  cracked  ice  and  water. 

Procedure:  Lubricate  the  stop  cock  of  the  dropping  funnel 
lightly  with  vaseline,  and  secure  it  in  place  with  a  rubber  band, 
so  that  it  may  not  slip  out  when  in  use  and  let  the  bromine  spurt 
out  over  the  ringers.  Secure  the  funnel  with  a  clamp  to  the  lamp 
stand  so  that  the  bottom  of  the  stem  is  about  one  inch  above  the 
ice  water.  Place  50  cc.  of  water  and  the  ammonium  hydroxide  in 
the  Erlenmeyer  flask,  float  the  latter  in  the  ice  water  and  insert 
the  stem  of  the  funnel  in  the  flask  so  that  it  is  kept  from  tipping 
over.  Pour  the  bromine  in  the  funnel.  Now  holding  the  bulb 
of  the  funnel  with  the  left  hand,  turn  the  stop  cock  with  the  right 
hand  to  let  one  drop  of  bromine  fall  into  the  flask;  quickly  close 
the  stop  cock  and  with  the  right  hand  grasp  the  flask  and  rotate 
its  contents.  Proceed  in  this  way  until  all  the  bromine  is  added, 
working  as  rapidly  as  possible  yet  avoiding  heating  the  flask  and 
causing  white  smoke,and  bromine  vapor  to  belch  out.  If,  when  all 
the  bromine  is  added,  the  solution  is  yellow  or  red,  add  ammonium 


STRONTIUM  HYDROXIDE  FROM  STRONTIUM  SULPHATE     167 

hydroxide,  a  drop  at  a  time,  until  the  color  disappears.  Pour 
the  solution  into  the  evaporating  dish  and  evaporate  it  to  dryness 
on  the  steam  bath  (see  Note  00  page  00).  Pulverize  the  dry  salt 
and  put  it  in  a  6-inch  cork  stoppered  test  tube. 

QUESTIONS 

1.  What  products  would  be  formed  if  bromine  were  added  to 
a  solution  of  sodium  hydroxide  instead  of  ammonium  hydroxide, 
if  the  solution  were  kept  cold?      —  if  it  were  heated? 

2.  Add  about  10  drops  of  bromine  to  10  cc.  of  a  cold  10  per- 
cent sodium  hydroxide  solution.     Add  this  gradually  to  a  solu- 
tion of  ammonium  hydroxide,  made  by  diluting  1  cc.  of  desk 
reagent  with  10  cc.  of  water.     Determine  what  gas  is  given  off. 

3.  What  fraction  of  the  entire  amount  of  ammonium  used  is 
lost  through  formation  of  nitrogen  gas  when  ammonium  bromide 
is  made  by  the  action  of  bromine  on  ammonium  hydroxide? 

4.  Why  cannot  hydrobromic  acid  be  prepared  from  potassium 
bromide  by  a  method  analogous  to  that  used  in  the  manufacture  of 
hydrochloric  acid? 

5.  Explain  why,  from  the  standpoint  of  economy,  the  method 
of  preparation  above  outlined  is  superior  to  the  direct  neutral- 
ization of  ammonia  with  hydrobromic  acid. 

PREPARATION  20 
STRONTIUM  HYDROXIDE  FROM  STRONTIUM  SULPHATE 

One  of  the  most  important  sources  of  strontium  is  the  mineral 
celestite,  SrSO4.  By  reduction  with  charcoal  this  can  be  con- 
verted into  strontium  sulphide, 

SrSO4  +  4C  =  SrS  +  4CO, 

and  the  strontium  sulphide  by  treatment  with  copper  oxide  and 
water  can  be  made  to  yield  strontium  hydroxide, 

SrS  +  CuQ  +  H2O  =  Sr(OH)2  +  CuS. 

Copper  oxide  is  in  the  ordinary  sense  insoluble;  nevertheless  in 
contact  with  water  it  does  yield  to  an  infinitesimal  extent,  first 
copper  hydroxide,  and  then  Cu++  ions, 

CuO  +  H20  ^  Cu(OH)2  ^±  Cu++  +  2OH- 


168  ALKALI  AND  ALKALINE  EARTH  METALS 

Therefore,  since  copper  sulphide  is  a  far  more  insoluble  substance 
than  copper  oxide,  it  follows  that  the  few  Cu++  ions  from  the 
latter  unite  with  the  S  ions  from  the  strontium  sulphide  to 
form  copper  sulphide,  which  precipitates  continuously,  while  the 
copper  oxide  continuously  gges  into  solution  to  resupply  Cu++ 
ions,  and  this  action  continues  until  either  the  copper  oxide  or  the 
strontium  sulphide  is  exhausted. 

Strontium  hydroxide  crystallizes  with  8  molecules  of  water, 
Sr(OH)2.8H2O.  It  is  very  soluble  in  hot  water,  but  sparingly 
soluble  in  cold  water. 

Materials:     celestite,  SrSO4,  61  grams  =  J  F.W. 

powdered  charcoal,  35  grams. 

copper  oxide,  48  grams. 
Apparatus:  gas  furnace. 

clay  crucible. 

large  mortar. 

8-inch  dish. 

500  cc.  flask. 

suction  filter. 

Procedure:  Grind  the  powdered  celestite  in  a  porcelain  mortar 
until  no  more  grit  is  felt  under  the  pestle.  Add  24  grams  of 
powdered  charcoal  and  continue  to  grind  with  the  pestle  until  the 
two  are  thoroughly  mixed.  Place  the  mixture  in  a  clay  crucible, 
pack  it  firmly,  and  cover  it  with  a  layer  of  powdered  charcoal  J 
inch  deep.  Cover  the  crucible  with  a  close-fitting  cover  and  heat 
it  in  a  gas  furnace  for  one  hour,  at  a  bright  red  heat.  After  the 
contents  of  the  crucible  have  cooled,  remove  the  layer  of  charcoal 
from  the  surface  and  bring  the  remainder,  after  crushing  it  to 
a  powder,  into  an  8-inch  porcelain  dish;  add  360  cc.  of  water, 
bring  the  mixture  to  a  boil,  and  while  it  is  boiling  add  copper 
oxide,  a  little  at  a  time,  until  all  of  the  soluble  sulphide  has  inter- 
acted with  it,  —  about  48  grams  in  all.  So  long  as  any  unchanged 
strontium  sulphide  is  present  the  solution  will  show  a  yellow  color, 
which  may  be  observed  by  letting  the  black  solid  settle  for  a  mo- 
ment, and  tnen  looking  through  the  upper  layers  of  the  clear  liquid 
at  the  background  of  the  white  porcelain  dish.  As  soon  as  the 
yellow  color  has  entirely  disappeared,  the  strontium  sulphide  has 
all  reacted.  Crystals  of  strontium  hydroxide  separate  rapidly 
from  this  solution  when  it  cools.  Hence  it  must  be  filtered  quickly 


STRONTIUM  CHLORIDE  FROM  STRONTIUM  SULPHATE      169 

in  order  to  avoid  having  the  crystals  form  in  the  filter  and  clog  it 
completely.  Heat  50  cc.  of  water  to  boiling  in  a  beaker,  and  keep 
it  at  this  temperature  until  it  is  required.  Add  hot  water  to  the 
dish  to  replace  any  lost  by  evaporation,  and  pour  (Note  2)  the  hot 
solution  through  a  large  ordinary  filter  (Note  4  (c)),  catching  the 
filtrate  in  a  500  cc.  flask,  and  allowing  the  main  part  of  the  residue 
to  remain  in  the  dish.  Add  the  50  cc.  of  hot  water  to  this  residue, 
stir  it  thoroughly,  heating  it  for  a  moment  over  the  flame,  and  then 
pour  solution  and  residue  into  the  filter  and  drain  out  all  of  the 
liquid.  Stopper  the  flask  to  exclude  the  air,  and  wrap  it  with  a 
towel,  so  that  the  solution  may  cool  slowly  and  larger  crystals 
may  be  formed.  Finally,  after  several  hours  cool  the  solution  with 
running  tap  water  and  then  collect  the  crystals  on  a  suction  filter. 
Drain  the  crystals  for  a  moment,  but  do  not  draw  too  much  air 
through  them,  as  they  retain  all  the  carbon  dioxide  it  contains. 
Wrap  the  product  in  paper  towels  and  leave  it  to  dry  over  night 
at  room  temperature.  (See  Note  9(6)  page  15.)  Put  the  product 
in  a  6-ounce  cork-stoppered  bottle. 

QUESTIONS 

1.  What  constituent  of  the  atmosphere  must  be  excluded  from 
the  solution  while  crystallizing  and  from  the  crystals  while  drying? 
How  would  it  contaminate  the  preparation? 

2.  A  sample  of  the  preparation  should  dissolve  nearly  clear  in 
hot  water.     What  will  surely  cause  a  slight  cloudiness? 

3.  How  could  strontium  chloride  be  prepared  from  strontium 
sulphide? 

4.  Give  some  other  method  by  which  strontium  hydroxide 
could  be  obtained  from  strontium  sulphide  without  the  use  of 
copper  oxide. 

5.  Starting  with  the  mineral  strontium  carbonate,  how  might 
strontium  hydroxide  be  prepared?     Strontium  oxide?     Strontium 
chloride? 

PREPARATION  21 
STRONTIUM  CHLORIDE  FROM  STRONTIUM  SULPHATE 

Strontium  chloride  might  be  prepared  by  treating  strontium 
sulphide,  the  intermediate  product  in  the  last  preparation,  with 
hydrochloric  acid,  but,  to  avoid  the  hydrogen  sulphide  nuisance, 
and,  furthermore,  to  show  that  strontium  sulphate  may  be  at- 


170  ALKALI  AND  ALKALINE  EARTH   METALS 

tacked  without  the  use  of  a  furnace,  we  shall  employ  quite  a  differ- 
ent method. 

The  method  consists  in  first  converting  the  sulphate  into  the 
carbonate  by  boiling  it  with  a  concentrated  solution  of  sodium 
carbonate,  and  then  dissolving  the  carbonate  in  hydrochloric 
acid,  thereby  yielding  a  solution  of  the  chloride.  The  conversion 
of  solid  strontium  sulphate  into  solid  strontium  carbonate  fur- 
nishes an  interesting  illustration  of  the  solubility  product  principle, 
for  the  solubility  of  these  two  salts  in  pure  water  is  as  follows : 


Solubility  in  grams 
per  liter 

Solubility  in  mols 
per  liter 

SrS04  
SrCO3... 

0.011 
0.0011 

0.0006 
0.00007 

Strontium  sulphate  would  dissolve  in  the  solution  of  sodium 
carbonate  in  the  same  manner  as  it  would  in  pure  water  until 
it  had  saturated  the  solution  and  its  solubility  product,  which  is 
equal  to  0.0006  X  0.0006,  was  reached,  but  for  the  fact  that  long 
before  this  could  occur  the  solution  would  be  supersaturated  with 
respect  to  strontium  carbonate,  whose  solubility  product  is  only 
equal  to  0.00007  X  0.00007.  Thus  strontium  carbonate  is  pre- 
cipitated continuously  as  strontium  sulphate  dissolves;  and  since 
the  solution  cannot  become  saturated  with  the  latter  so  long  as 
there  is  a  large  excess  of  carbonate  ions  present,  the  solid  salt 
finally  remaining  will  consist  entirely  of  strontium  carbonate, 
provided  a  sufficient  amount  of  sodium  carbonate  were  employed. 
The  reaction  which  takes  place  is,  however,  reversible,  SrSCX  + 
NaaCOs  ^  SrCOs  +  Na2SO4,  and,  if  strontium  carbonate  were 
boiled  with  a  solution  of  sodium  sulphate,  the  solid  would  change 
into  sulphate  until  carbonate  ions  had  accumulated  in  the  solution 
to  such  an  extent  as  to  make  the  concentration  ratio  [COs  ]/ 
[SO4 — ]  =  71!-.  When  this  ratio  prevails,  both  solids  are  in 
equilibrium  with  the  solution  and  no  change  takes  place  in  either 
direction. 

Materials:     celestite,  SrS04,  61  grams  =  J  F.W. 
anhydrous  Na2C03,  73  grams, 
chlorine  water. 
6-normal  HC1,  111  cc.  =  f  F.W. 


STRONTIUM  CHLORIDE  FROM  STRONTIUM  SULPHATE     171 

Apparatus:  8-inch  porcelain  dish. 
600  cc.  tall  beaker. 

Procedure:  Take  the  powdered  celestite.  Grind  it  in  a  mortar 
until  it  is  so  fine  that  it  no  longer  feels  gritty  under  the  pestle. 
Cover  it  in  an  8-inch  dish  with  360  cc.  of  water,  add  the  anhydrous 
sodium  carbonate,  and  boil  the  mixture  for  30  minutes,  stirring  it 
constantly  at  first.  Transfer  the  solution  and  solid  to  a  tall, 
narrow  beaker,  using  100  cc.  of  fresh  water  in  rinsing  out  the  last 
of  the  residue,  and  let  the  solid  matter  settle  for  5  minutes.  De- 
cant off  the  liquid,  which  is  still  somewhat  cloudy,  but  from  which 
the  essential  part  of  the  solid  has  settled,  and  wash  the  residue 
three  times  by  decantation  with  400-500  cc.  of  water  (see  Note  5 
(6),  first  paragraph).  The  residue  is  now  sufficiently  free  from 
soluble  sodium  sulphate.  Transfer  about  TV  of  the  moist  stron- 
tium carbonate  to  another  beaker,  to  be  used  in  a  later  part  of  the 
process.  To  the  remaining  T\  add  50  cc.  of  hot  water,  and  then 
add  hydrochloric  acid,  drop  by  drop,  while  keeping  the  mixture 
at  the  boiling  temperature,  until  the  further  addition  of  a  drop  of 
acid  produces  no  more  effervescence.  This  solution  now  contains 
a  slight  excess  of  acid,  and  probably  a  trace  of  iron  chloride  as 
impurity.  Add  a  few  drops  of  chlorine  water  to  oxidize  any 
ferrous  to  ferric  salt,  then  add  the  remaining  tenth  of  the  strontium 
carbonate  and  boil  the  mixture  for  5  minutes.  The  solution 
should  now  be  perfectly  neutral  to  litmus  in  which  case  all  iron 
will  be  precipitated  as  Fe(OH)3.  If  it  is  acid,  it  shows  that  the 
hydrochloric  acid  was  added  carelessly  and  that  there  was  thus 
more  than  could  be  neutralized  by  the  strontium  carbonate.  Filter 
the  perfectly  neutral  solution,  and  evaporate  the  filtrate  until  a 
faint  scum  forms  on  removing  the  solution  from  the  flame  and 
blowing  vigorously  across  the  surface.  Allow  the  solution  to  cool, 
but  stir  occasionally  in  order  to  obtain  a  uniform  crystal  meal 
rather  than  a  cake  of  crystals.  Finally,  drain  the  crystals  on  a 
suction  filter  (Note  4  (6));  evaporate  the  mother  liquor  to  crys- 
tallation  exactly  as  at  first,  and  if  the  second  crop  of  crystals  is 
pure  white,  add  it  to  the  first  crop.  Wrap  the  crystals  of  SrCl2.- 
6H2O  in  paper  towels  and  leave  them  overnight  to  dry.  (Note  9(6) , 
page  15.)  These  crystals  are  efflorescent,  hence,  as  soon  as  the 
paper  package  is  unwrapped,  place  them  in  a  2-ounce  cork-stop- 
pered bottle. 


172  ALKALI  AND  ALKALINE  EARTH   METALS 

QUESTIONS 

1.  Explain  why  strontium  carbonate,  which  is  less  soluble  in 
pure  water  than  strontium  sulphate,  should  dissolve  readily  in 
dilute  acids,  while  the  latter  salt  will  dissolve  scarcely  any  more 
in  acids  than  in  pure  water,  f 

2.  If  a  small  quantity  of  a  solution  of  strontium  chloride  were 
added  to  a  solution  containing  equi-molal  quantities  of  sodium 
carbonate  and  sodium  sulphate,  what  would  be  the  precipitate 
formed? 

PREPARATION  22 

BARIUM  OXIDE  AND  BARIUM  HYDROXIDE  FROM 
BARIUM  CARBONATE 

The  commercial  method  of  preparing  calcium  oxide  (quicklime) 
consists  in  heating  calcium  carbonate  (limestone)  in  lime  kilns. 
Barium  oxide  might  be  made  from  barium  carbonate  according 
to  the  same  principle,  except  for  the  fact  that  the  temperature 
required  for  the  decomposition  of  barium  carbonate  is  so  high  as 
to  make  such  a  method  almost  impracticable.  This  greater 
stability  of  the  barium  salt  is  an  illustration  of  the  fact  that  barium 
oxide  is  even  more  strongly  basic  than  calcium  oxide.  The  reac- 
tion, BaCOs  ^  BaO  +  C02,  is  to  some  extent  reversible,  and  in 
common  with  other  reversible  reactions  it  may  be  made  to  pro- 
gress in  one  direction  or  the  other  by  suitably  altering  the  con- 
centration of  the  substances  present  in  the  reacting  system.  Of 
the  three  substances  involved  in  this  reaction  the  only  one  which 
can  be  removed  during  the  course  of  the  reaction  is  the  carbon 
dioxide.  It  is,  however,  not  enough  to  let  it  merely  pass  off  as  a 
gas,  because  there  is  already  present  in  the  atmosphere  in  the 
furnace  enough  carbon  dioxide  to  force  the  reaction  towards  the 
left.  The  carbon  dioxide  must  be  chemically  removed.  This  is 
accomplished  by  mixing  powdered  charcoal  with  the  barium  car- 
bonate, for  carbon  reacts  with  carbon  dioxide  at  a  white  heat  and 
gives  carbon  monoxide.  In  the"  following  procedure,  in  addition 
to  the  charcoal,  a  little  rosin  is  mixed  with  the  charge.  On  heating, 
the  rosin  decomposes  and  a  deposit  of  soot  is  formed,  which  in  this 
way  becomes  very  intimately  mixed  with  the  charge. 

The  barium  oxide  obtained  in  this  way  is  not  pure,  but  contains 
particles  of  charcoal  as  well  as  impurities  coming  from  the  mineral. 


BARIUM  OXIDE  AND  BARIUM  HYDROXIDE  173 

It  is,  however,  very  suitable  for  the  manufacture  of  barium  hy- 
droxide, into  which  it  is  converted  by  treatment  with  water. 
Barium  hydroxide  is  extremely  soluble  in  hot  water,  but  sparingly 
so  in  cold  water,  from  which  it  separates  in  flake-like  crystals  of 
the  composition  Ba(OH)2.8H2O. 

Materials:  barium  carbonate,  BaCO3,  99  grams  =  J  F.W.;  if 
the  mineral  witherite  is  used  it  should  be  very 
finely  powdered;  the  artificially  prepared  mate- 
rial will  react  more  readily. 

powdered  charcoal,  25  grams. 

rosin,  5  grams. 
Apparatus:  gas  furnace. 

clay  or  graphite  crucible. 

8-inch  porcelain  dish. 

5-inch  filter  funnel  and  plain  filter. 

500  cc.  flask. 

Procedure:  Mix  the  finely  powdered  barium  carbonate  with 
10  grams  of  powdered  charcoal  and  5  grams  of  powdered  rosin. 
After  mixing  the  whole  mass  very  thoroughly  in  a  mortar,  place 
it  in  the  crucible,  press  it  down  firmly,  and  cover  it  with  a  layer 
of  charcoal  at  least  \  inch  deep.  Place  a  well-fitting  cover  on  the 
crucible  and  heat  the  whole  for  one  hour  to  as  high  a  temperature 
as  possible  in  the  gas  furnace.  After  the  crucible  has  cooled, 
remove  the  top  layer  of  charcoal,  place  the  barium  oxide  in  the 
porcelain  dish,  and  very  cautiously  add  a  few  drops  of  water, 
noting  the  very  violent  reaction.  Finally  add  400  cc.  of  water, 
and  heat  the  mixture  in  the  dish  to  boiling;  pour  the  solution 
through  a  large,  ordinary  filter  (Note  4  (c)),  letting  the  clear  liquid 
run  directly  into  a  500  cc.  flask.  Rinse  the  residue  in  the  dish  with 
75  cc.  more  of  boiling  water,  and  pour  this  upon  the  filter  after 
the  first  portion  has  nearly  all  run  through.  Stopper  the  flask 
and  allow  the  solution  to  cool  slowly  to  room  temperature;  finally, 
cool  it  nearly  or  quite  to  0°;  collect  the  crystals  on  a  suction  filter; 
wrap  the  product  in  paper  towels  and  leave  it  overnight  to  dry 
at  room  temperature.  Put  the  product  in  a  6-ounce  cork-stop- 
pered bottle. 


174  ALKALI  AND  ALKALINE  EARTH   METALS 

QUESTIONS 

1.  How  does  barium  hydroxide  become  contaminated  by  ex- 
posure to  the  air?     Why  might  this  product  be  dried  with  less 
contamination  by  exposure  to  the  air  out  of  doors  than  to  the  air 
of  the  laboratory? 

2.  The  mineral  witherite  often  contains  barium  sulphate  as 
an  impurity.     State  what  changes  this  substance  would  undergo 
during  the  above  process,  and  what  resulting  substance  would 
be  present  to  a  small  extent  in  the  final  crystallized  product  and 
to  a  greater  extent  in  the  mother  liquor.     (Compare  preparation 
of  strontium  hydroxide  from  strontium  sulphate.)     Test  both 
crystals  and  mother  liquor  for  this  substance.     How?     The  final 
product  can  be  quite  satisfactorily  purified  from  it  by  one  or  two 
recrystallizations.     How  might  it  be  removed  chemically? 

3.  Devise   a   method  for   preparing   barium   hydroxide   from 
barium  carbonate  by  which  the  use  of  a  furnace  may  be  avoided. 
Suggestion:  Make  use  of  the  difference  in  solubility  of  barium 
chloride  and  barium  hydroxide. 

Experiments 

Very  few  of  the  chemical  properties  of  the  non-metals  are  dis- 
played except  in  conjunction  with  the  metals;  in  our  study  of  the 
non-metals  in  the  preceding  chapters,  therefore,  we  already  have 
had  revealed  to  us  many  of  the  chemical  properties  of  the  metals. 
Preceding  experiments  which  involve  the  alkali  and  alkaline  earth 
metals,  and  which  should  now  be  reviewed  are : 

Chapter  II,  Experiments  3,  4,  5,  6,  7,  11,  13. 
Chapter  III,  Experiments  2,  5,  6,  7,  8,  9,  11,  12. 
Chapter  IV,  Experiments  4,  6,  25. 

1.  Place  a  few  small  lumps  of  marble  (pure  calcium  carbonate) 
in  a  small  porcelain  crucible.     Cover  the  crucible  in  order  to  keep 
in  the  heat,  and  heat  it  strongly  for  20  minutes  with  a  Bunsen 
flame.     When  the  product  has  cooled,  wet  each  lump  with  a 
single  drop  or  two  of  water  and  wait  a  few  minutes,  if  necessary, 
to  observe  the  effect.     Then  wet  the  product  with  somewhat 
more  water,  and  test  the  reaction  of  the  moist  mass  towards  litmus. 

2.  Place  a  few  grams  of  magnesium  carbonate  in  a  3-inch  dish 
and  heat  it  rather  moderately,  testing  to  see  if  carbon  dioxide  is 


EXPERIMENTS  175 

being  expelled.  From  time  to  time  test  the  residue  for  carbonate 
by  removing  a  little  from  the  dish,  thoroughly  wetting  it  with 
about  5  cc.  of  water  in  a  test  tube,  adding  acid,  and  watching  for 
effervescence. 

3.  Burn  a  strip  of  magnesium  ribbon,  held  with  iron  pincers, 
and  let  the  ash  fall  in  a  porcelain  dish.     Wet  the  magnesium 
oxide  with  a  single  drop  of  water  and  place  the  moist  mass  on  a 
strip  of  red  litmus  paper.     Note  the  rapidity  and  intensity  with 
which  the  litmus  is  turned  blue. 

4.  To  some  magnesium  chloride  solution,  add  (a)  some  am- 
monium hydroxide;    (6)  some  ammonium  chloride  and  then  some 
ammonium  hydroxide.     Observe  in  each  case  whether  magnesium 
hydroxide  is  precipitated. 

5.  Dip  a  clean  platinum  wire  in  solutions  of  such  of  the  chlorides 
of  the  alkali  and  alkaline  earth  metals  as  are  at  hand,  and  observe 
the  color  imparted  to  the  Bunsen  flame  when  the  wire  is  inserted 
into  the  lower  part  of  the  flame. 

AMMONIUM  COMPOUNDS 

When  ammonium  hydroxide  dissociates  electrolytically  it  yields 
the  ion  NH4+.  The  group  of  atoms  NH4,  which  is  often  spoken 
of  as  the  ammonium  radical,  resembles  in  many  respects  the  atom 
of  sodium  or  potassium.  Like  the  latter,  it  can  form  a  monovalent 
positive  ion,  or  it  can  form  compounds  with  acid  radicals,  for 
example,  NH4C1,  (NH4)2SO4;  but  unlike  sodium  and  potassium, 
it  cannot  exist  in  the  uncombined  state. 

6.  (a)  Place  a  mixture  of  dry  ammonium  chloride  and  calcium 
hydroxide  in  a  dry  test  tube  and  heat  gently.     Test  the  evolved 
gas  for  ammonia. 

(6)  Add  a  solution  of  a  strong  base  to  a  solution  of  any  ammo- 
nium salt,  warm,  and  notice  whether  the  odor  of  ammonia  is  appar- 
ent. Show  that  this  experiment  furnishes  an  instance  of  the 
displacement  of  a  weak  base  from  its  salt  by  means  of  a  strong 
base. 

7.  GASEOUS  DISSOCIATION  OF  AMMONIUM  CHLOKIDE. —  Like  all 
ammonium  salts,  ammonium  chloride  can  be  volatilized  by  apply- 
ing heat.     Some  of  the  other  salts  are  permanently  decomposed 
by  the  process,  but  the  vapor  of  ammonium  chloride  can  be 
condensed  again  to  the  same  solid  substance.     In  the  vapor  con- 


176  ALKALI  AND  ALKALINE  EARTH  METALS 

dition,  however,  the  salt  is  highly  dissociated  into  two  gaseous 
compounds. 

Place  2  grams  of  ammonium  chloride  in  the  middle  of  an  8-inch 
piece  of  hard  glass  tube  and  on  either  side  place  loose  plugs  of 
asbestos.  Outside  the  asbestos  plugs  place  moistened  strips  of 
both  red  and  blue  litmus  paper  at  both  ends.  Support  the 
tube  in  a  slightly  inclined  position  by  means  of  a  clamp,  and 
heat  the  section  containing  the  salt,  using^  a  flame  spreader. 
Observe  the  changes  in  color  of  the  litmus. 

Remembering  that  gases  diffuse  with  velocities  inversely  pro- 
portional to  the  square  roots  of  their  densities,  account  for  the 
effects  observed. 

8.  HYDROLYSIS  OF  AMMONIUM  SALTS. —  Boil  for  some  time  a 
solution  of  ammonium  sulphate  to  which  has  been  added  a  few 
drops  of  blue  litmus  solution.  Pass  the  vapors  into  a  flask  of 
water  containing  a  few  drops  of  red  litmus. 

Explain  why  boiling  increases  the  extent  of  the  hydrolysis  of 
the  salt. 

GENERAL  QUESTIONS  V 
ALKALI  AND  ALKALINE  EARTH  METALS 

1.  Make  a  table  of  the  elements  of  Group  I,  Family  A,  includ- 
ing lithium  and  sodium  giving  in  succeeding  columns:     1,  the 
symbol  of  the  element;   2,  its  valence  in  its  compounds;    3,  the 
formula  of  the  oxide;    4,  the  formula  of  the  hydroxide;    5,  the 
solubility  of  the  hydroxide  in  grams  per  100  grams  of  water  at 
25°;   6,  the  formula  of  the  sulphate;   7,  the  solubility  of  the  sul- 
phate at  25°;   8,  the  formula  of  the  carbonate;   9,  the  solubility 
of  the  carbonate  at  25°;    10,  the  formula  of  the  chloride;   11,  the 
solubility  of  the  chloride  at  25°. 

2.  Make  a  similar  table  of  the  elements  of  Group  II,  Family  A, 
including  magnesium. 

3.  Give  the  same  information  for  ammonium.     Discuss  the 
difference  between  ammonium  and  ammonia. 

4.  Make  a  list  of  the  percent  of  ionization  of  the  hydroxides 
of  the  alkali  metals,  of  ammonium,  and  of  the  alkaline  earth 
metals  in  0.1  normal  solution  if  the  substance  is  soluble  to  that 
extent.     Give  figures  for  the  hydroxyl  ion  concentration  in  0.1 


GENERAL  QUESTIONS  177 

normal  NH4OH  solution  and  in  saturated  solutions  of  Ca(OH)2 
and  Mg(OH)2.  Describe  and  discuss  the  results  of  Experiment 
4  in  the  light  of  these  figures. 

5.  An  oxy-salt,  such  as  CaC03  (=  CaO.CO2),  can  be  broken 
up  by  a  sufficiently  high  heat  into  a  basic  oxide  and  an  acid  oxide  — 
for  example,  CaO.CO2  — >  CaO  -f  C02.     The  higher  the  tempera- 
ture necessary  to  accomplish  this,  the  greater  is  the  chemical 
affinity  between  the  two  oxides,  that  is,  the  more  strongly  basic 
and  acidic,  respectively,  are  these  two  components;  and  therefore 
in  a  series  of  salts,  all  containing  the  same  acidic  oxide  —  for 
example,  CaCO3,  SrCO3,  BaCO3  —  the  greater  the  stability  of  the 
salt,  the  stronger  is  the  basic  oxide.     Compare  the  approximate 
temperatures  at  which  the  alkaline  earth  carbonates  are  decom- 
posed, and  list  the  alkaline  earth  oxides  in  the  order  of  their 
basic  strength. 

The  carbonates  of  the  alkali  metals  are  practically  undecom- 
posable  by  heat  alone.  Compare  the  basic  strength  of  the  alkali 
metal  oxides  as  a  family  with  that  of  the  alkaline  earth  oxides. 

6.  More  precise  information  as  to  the  relative  basic  strength 
may  be  given  from  the  molal  heats  of  formation  in  a  series  of 
oxy-salts    like    the    carbonates.     In   the   thermochemical   tables 
sometimes  the  figure  that  we  want  is  given,  namely  the  heat  of 
formation  of  the  salt  from  the  metal  oxide  and  the  non-metal  oxide, 
for    example,    MgO  +  C02  -»  MgCO3  +  28,850    calories.     But 
more  often  we  will  find  only  the  heat  of  formation  from  the  ele- 
ments,   for    example,    Mg  +  C  +  |O2  — >  MgCO3  +  269,900    cal- 
ories.    In  this  case  we  must  look  up  the  heat  of  formation  of 
magnesium  oxide,  Mg  +  |02  — >  MgO  +  143,400  and  the  heat  of 
formation  of  carbon  dioxide,   C  +  O2  — >  C02  +  97,650  calories, 
and  subtract  the  sum  of  these  quantities:    269,900  —  143,400  — 
97,650  =  28,850"  calories. 

Find  in  the  thermochemical  tables  the  molal  heat  of  formation 
of  the  carbonates  of  sodium,  potassium,  calcium  and  barium,  from 
the  metal  oxide  and  carbon  dioxide  and  draw  conclusions  as  to  the 
relative  basic  strength  of  the  basic  oxides. 

7.  State  the  colors  imparted  to  a  Bunsen  flame  by  vaporized 
salts  (chlorides)  of  the  alkali  and  alkaline  earth  metals. 

8.  Give  formulas  of  the  peroxides  of  sodium,   and  barium. 
How  do  these  substances  react  with  cold  dilute  acids?     What  is 


178  ALKALI  AND  ALKALINE  EARTH  METALS 

the  formula  of  the  peroxide  of  potassium?  Compare  the  action 
of  the  oxides  Na20,  K20,  and  BaO  in  water,  with  that  of  the  per- 
oxides. What  is  the  valence  of  the  metal  in  each  of  the  peroxides, 
and  how  do  you  account  for  the  amount  of  oxygen  over  that  con- 
tained in  the  oxide? 

***<«  • 


CHAPTER  VI 

ELEMENTS   OF  GROUP  III  OF  THE 
PERIODIC   SYSTEM 

Boron  and  aluminum,  the  first  two  members  of  this  group,  are 
the  only  ones  which  are  classed  among  the  common  elements. 
On  this  account,  and  also  because  the  difference  in  properties 
between  Family  A  and  Family  B  is  far  less  marked  than  in  Groups  I 
and  II,  the  whole  group  is  taken  up  under  one  heading. 

The  characteristics  of  this  group  are  that  the  elements  possess 
a  valence  of  three,  and  that  the  oxides,  M2O3,  have  but  a  weakly 
developed  basic  character.  Boron,  in  fact,  shows  practically 
no  base-forming  properties,  but  forms  rather  the  weak  boric  acid. 
The  oxide  of  aluminum  displays  both  basic  and  acidic  properties; 
that  is,  it  is  amphoteric.  The  remaining  elements  are  more  dis- 
tinctly base-forming  than  aluminum,  without,  however,  approach- 
ing in  any  way  the  alkaline  earth  metals  in  this  respect. 

PREPARATION  23 
BORIC  ACID 

In  this  preparation,  borax,  the  sodium  salt  of  tetraboric  acid, 
is  chosen  as  the  source  of  boron.  Although  boron  is  decidedly  a 
non-metal,  still  its  acid-forming  characteristics  are  not  highly 
developed  and  its  acids  are  readily  displaced  by  strong  acids  from 
solutions  of  their  salts.  Thus  tetraboric  acid,  H2B4O7  would  be 
set  free  from  borax  by  hydrochloric  acid,  but  the  acid  which 
actually  crystallizes  is  the  more  highly  hydrated  orthoboric  acid 
H3B03. 

Materials:    borax,  Na2B4O7.10H2O,  96  grams  =  |  F.W. 

12-normal  HC1. 

methyl  orange  solution. 
Apparatus:  500  cc.  beaker. 

suction  filter. 

Procedure:  Dissolve  the  borax  in  300  cc.  of  boiling  water.  Add 
a  few  drops  of  methyl  orange  solution  (a  dye  which  is  yellow  in 

179 


180  ELEMENTS  OF  G&OUP  III 

neutral  or  alkaline  solution  but  is  pink  in  distinctly  acid  solution) 
and  add  12-n  HC1  until  the  color  of  the  dye  has  changed  through 
an  orange  to  a  distinct  pink  and  addition  of  1  cc.  more  of  the  acid 
does  not  increase  the  pink  tone.  Let  the  solution  cool  to  15°  or 
below,  drain  the  crystals  on  a- suction  filter.  If  the  filtrate  is  not 
distinctly  pink  (showing  it  tb  be  acid)  add  enough  12-n  HC1  to 
make  it  so,  shake  vigorously,  and  add  any  crystals  thus  obtained 
to  the  main  crop.  Dissolve  the  crystals  in  300  cc.  of  boiling  water, 
filter  if  not  clear,  crystallize  by  slow  cooling  with  occasional 
stirring  if  crystals  cake  together  too  much.  Collect  the  crystals 
and  let  them  dry  at  room  temperature. 

Note.  —  The  above  procedure  yields  fine  granular  crystals. 
Lustrous  flaky  crystals  can  be  obtained  if  a  little  grease  is  present 
in  the  crystallizing  solution.  To  this  end  dissolve  1  gram  of  soap 
shavings  in  300  cc.  of  boiling  water  and  use  this  solution  to  dis- 
solve the  boric  acid  for  the  recrystallization.  This  procedure  is 
not  recommended  if  purity  of  product  is  required  rather  than  an 
attractive  appearance. 


QUESTIONS 

1.  Explain  the   relations   between   normal  boric   acid,   meta- 
boric  acid,  tetra Doric  acid,  and  boric  anhydride.     Experiment: 
Place  a  few  grams  of  boric  acid  on  a  watch  glass  upon  the  steam 
table  (100-110°)  and  leave  for  J  hour.     What  is  formed?    What 
would  be  formed  if  the  acid  were  heated  to  140°?     Suspend  a 
little  of  the  acid  in  a  loop  of  platinum  wire,  and  heat  in  the  Bunsen 
flame.     What  is  formed? 

2.  Experiment:  Place  a  few  grains  of  boric  acid  in  a  small 
porcelain  dish,  cover  it  with  5  cc.  of  alcohol,  set  fire  to  it,  and  ob- 
serve the  color  of  the  edges  of  the  flame,  especially  when  stirring 
and  when  the  alcohol  is  almost  burned  out.     Repeat,  using  borax 
instead  of  the  boric  acid,  and  again,  using  borax  moistened  with 
concentrated  sulphuric  acid. 

What  causes  the  green  color  of  the  flame,  and  why  is  it  not 
observed  with  borax  alone? 

Repeat  if  necessary  the  last  part  of  the  preceding  experiment, 
noticing  the  color  imparted  to  the  flame  while  the  orthoboric 
acid  is  first  melting,  and  again  when  a  clear  bead  of  boric  anhy- 
dride is  obtained. 


ALUM   FROM   CRYOLITE  181 

What  conclusions  can  you  make  from  these  experiments  regard- 
ing the  volatility  of  boric  acid  and  of  boric  anhydride? 

3.  What  effect  has  a  solution  of  borax  upon  litmus?     Explain 
what  is  thus  shown  regarding  the  strength  of  boric  or  tetraboric 
acid.     Explain  why  litmus  will  not  be  turned  a  bright  red  until 
more  than  two  moles  of  HC1  have  been  added  to  one  mole  of  borax. 

4.  How  can  boron  chloride  be  prepared?     How  does  this  sub- 
stance behave  when  treated  with  water?     How  would  it  behave 
if  boron  were  a  strongly  metallic  element? 

PREPARATION  24 

ALUM  FROM  CRYOLITE 

(BY-PRODUCT:  SODIUM  CARBONATE) 

A  characteristic  series  of  compounds  of  the  trivalent  metals 
are  the  alums,  of  which  potassium  aluminum  sulphate,  or  common 
alum,  K2SO4.A12(S04)3.24H20,  is  typical.  These  compounds  are 
particularly  interesting  from  the  readiness  with  which  they  can 
be  produced  in  large  and  beautiful  crystals,  and  from  the  fact 
that  any  of  the  univalent  alkali  metals  may  take  the  place  of 
potassium,  or  any  one  of  a  large  number  of  the  trivalent  metals 
may  take  the  place  of  aluminum  in  the  common  alum  without 
altering  the  form  of  the  crystals  produced. 

Of  the  many  raw  materials  which  might  serve  as  the  source  of 
aluminum,  cryolite,  the  double  fluoride  of  sodium  and  aluminum, 
3NaF.AlF3  or  Na3(AlF6),  has  been  chosen,  partly  because  it 
actually  serves  as  an  important  source  of  aluminum  compounds, 
and  partly  because  its  decomposition  illustrates  important  chemi- 
cal reactions  and  manipulations. 

This  mineral,  although  itself  insoluble  in  water,  will,  if  boiled 
with  milk  of  lime,  (a  suspension  of  Ca(OH)j),  undergo  a  meta- 
thesis, with  the  formation  of  insoluble  calcium  fluoride,  and  a  sol- 
uble salt  of  aluminum,  in  which  this  metal  plays  the  part  of  an 
acid-forming  element, 

3NaF.AlF3  +  3Ca(QH)2  =  3CaF2  -f  Na3AlO3  +  3H2O. 

By  removing  the  insoluble  residue  from  the  liquid,  a  separation 
of  the  aluminum  from  the  fluorine  is  accomplished;  but  on  the 
laboratory  scale  this  separation  is  difficult  to  carry  out  on  account 
of  the  colloidal  nature  of  the  residue.  If  filtration  were  resorted 


182  ELEMENTS  OF  GROUP  III 

to,  the  pores  of  the  filter  would  be  immediately  clogged  with  the 
gelatinous  precipitate,  so  that  the  liquid  would  run  so  slowly, 
even  with  suction,  that  an  undue  length  of  time  would  be  spent. 
Therefore  the  method  which  will  be  employed  is  that  of  sedimen- 
tation after  stirring  up  with  a iarge  amount  of  water.  It  must  be 
borne  in  mind,  however,  that  this  device  is  adopted  only  to  meet 
the  requirements  of  laboratory  practice,  for  on  a  commercial 
scale  the  expense  of  evaporating  the  large  amount  of  water  to 
obtain  soda,  one  of  the  by-products  of  the  process,  would 
be  prohibitive.  The  separated  sludge,  consisting  of  calcium 
fluoride  and  all  excess  of  calcium  hydroxide,  is  of  no  great  value, 
and  will  be  discarded,  although  it  could  be  used  as  a  source  of 
fluorine  compounds. 

From  the  clear  solution  of  sodium  aluminate  the  aluminum  is 
precipitated  by  displacement  of  the  weak  aluminic  acid  from  the 
salt  by  the  action  of  the  stronger  carbonic  acid, 

2Na3A103  +  3H2O  +  3CO2  =  3Na2CO3  +  2H3A103. 

This  precipitated  aluminic  acid,  H3A1O3  (or,  as  more  frequently 
named,  aluminum  hydroxide,  A1(OH)3),  is  also  a  very  gelatinous 
substance,  and  can  likewise  only  be  separated,  within  a  reasonable 
length  of  time,  by  means  of  sedimentation. 

The  aluminum  hydroxide  is  treated  with  the  calculated  amount 
of  sulphuric  acid  whereby  the  soluble  salt,  aluminum  sulphate, 
is  obtained.  To  this  solution  is  added  the  calculated  amount  of 
potassium  sulphate,  and  then  the  alum  is  allowed  to  crystallize. 

The  ideal  conditions  for  obtaining  large,  clear  crystals  —  which 
constitutes  the  beauty  of  this  as  a  laboratory  preparation  —  are, 
that  a  solution  which  is  just  saturated  with  alum  may  be  slowly 
concentrated  by  spontaneous  evaporation  at  a  nearly  constant 
temperature.  Such  conditions  are  found  in  industrial  works, 
where  the  evaporation  of  the  solution  in  large  vats  yields  beautiful 
crystals,  often  of  enormous  size;  but  these  necessary  conditions 
are  almost  impossible  to  realize  in  a  small  laboratory  preparation, 
and  another  method  is  adopted  to  give  more  rapidly  and  more 
surely  the  desired  results. 

From  the  accompanying  table  it  is  seen  that  the  solubility 
increases  rapidly  with  the  temperature.  If  a  solution  is  saturated 
at  35°  and  then  cooled  to  15-20°  about  one-half  of  the  alum  will 
separate  out,  but  ordinarily  in  the  form  of  a  mass  of  very  minute 


ALUM  FROM  CRYOLITE 


183 


100  grams  of  water  dissolve  the  given  number  of  grams  of 
K2SO4.A12(SO4)3.24H2O. 


Temp 

0° 

5° 

10° 

15° 

20° 

25° 

80° 

40° 

50° 

60° 

70° 

Grams  

5.6 

6.6 

7.6 

9.6 

11.4 

14.1 

16.6 

24 

36 

57 

110 

crystals.  In  order  to  obtain  large  crystals  during  the  cooling, 
three  precautions  are  necessary:  (1)  A  few  small  crystals  must  be 
added  to  serve  as  nuclei  for  the  crystallization  before  setting  the 
solution  to  cool.  (2)  Dust  must  be  excluded,  since  dust  particles 
might  serve  as  nuclei  for  the  formation  of  a  great  number  of  little 
crystals.  (3)  The  cooling  must  take  place  very  slowly  in  order 
that  the  crystal  faces  may  be  built  up  uniformly.  This  can  be 
accomplished  if  the  crystallizing  dish  is  insulated  by  being  covered 
with  a  watch  glass  and  wrapped  with  a  towel.  If  the  solution 
set  to  crystallize  is  saturated  at  a  much  higher  temperature  than 
35°,  it  will  be  found  that  the  crystallization  will  proceed  so  rapidly 
that  it  will  be  quite  impossible,  even  if  the  above  precautions  are 
all  observed,  to  obtain  good  crystals. 
Materials:  cryolite,  Na3AlF6,  53  grams  =  J  F.W. 

quicklime,  CaO,  48  grams. 

36-normal  H2SO4,  32  cc. 

potassium  sulphate,  K2S04,  22  grams  =  J  F.W. 
Apparatus:  500  cc.  casserole. 

2-liter  common  bottle. 

8-inch  evaporating  dish. 

6-inch  crystallizing  dish  and  watch  glass  or  plate 
to  cover  it. 

Procedure:  Slake  the  quicklime  and  then  mix  in  a  large  casserole 
with  250  cc.  of  water  to  make  milk  of  lime.  Stir  into  this  53  grams 
of  finely  powdered  cryolite,  and  while  constantly  stirring,  heat  to 
boiling.  Continue  boiling  for  an  hour,  adding  water  to  replace 
that  lost  by  evaporation  and  stirring  sufficiently  to  avoid  spatter- 
ing. At  the  end  of  that  time  add  a  part  of  1J  liters  of  boiling  water 
to  the  thick  mass  in  the  dish,  and  transfer  it,  together  with  the  rest 
of  the  hot  water,  to  a  tall,  wide-mouth,  2-liter  bottle,  and  set  aside 
to  settle  until  the  next  period.  The  sludge  should  have  settled 
so  as  to  occupy  not  more  than  one-fifth  of  the  volume  of  the  liquid. 


184  ELEMENTS  OF  GROUP  III 

Siphon  off  as  much  clear  liquid  as  possible  without  drawing  over 
any  of  the  precipitate.  Then  add  1J  liters  more  of  hot  water  to 
the  bottle,  stir  and  again  let  settle,  and  draw  off  the  clear  liquor. 
The  residue  in  the  bottle  may  be  thrown  away.  Combine  all 
the  solution  in  a  large  bottle,  an*!  pass  in  carbon  dioxide  (Note  13  (a) , 
page  19),  until  all  of  the  aluminum  is  precipitated.  Test  to  see 
if  this  is  accomplished  at  the  end  of  J  hour  by  stopping  the  carbon 
dioxide  stream  and  letting  the  precipitate  settle  enough  to  pour 
off  a  little  clear  liquor  into  a  beaker.  Pass  carbon  dioxide  into 
this  for  a  few  minutes;  if  no  precipitation  occurs  it  shows  that 
all  of  the  alumina  has  already  fallen  out  of  the  solution.  If  a 
precipitate  does  appear,  the  treatment  of  the  entire  solution  with 
carbon  dioxide  must  be  continued  until  all  the  alumina  is  thrown 
out.  Then  let  the  precipitate  settle  until  it  occupies  less  than 
one-sixth  of  the  entire  volume.  Siphon  off  the  clear  liquid  and 
evaporate  it  to  dryness  in  a  porcelain  dish.  Powder  the  sodium 
carbonate  so  obtained,  and  preserve  it  in  a  cork-stoppered  test 
tube.  Stir  up  the  precipitate  left  in  the  bottle  with  1J  liters  of 
hot  water;  let  settle,  and  siphon  off  and  discard  the  clear  liquid, 
since  it  will  not  contain  sufficient  sodium  carbonate  to  pay  for  its 
evaporation.  To  the  suspension  of  aluminum  hydroxide  left  in 
the  bottle  add  32  cc.  of  36-n  H2S04,  and  warm,  if  necessary,  to 
effect  complete  solution.  Add  22  grams  of  potassium  sulphate, 
and  warm  until  dissolved.  The  solution  should  now  be  perfectly 
clear;  if  not,  filter.  If  the  volume  exceeds  400  cc.  evaporate  to 
that  bulk,  and  while  still  hot  transfer  it  to  the  crystallizing  dish 
(an  8-inch  porcelain  dish  will  answer).  When  cooled  to  55-50°, 
drop  8  to  10  very  small  alum  crystals  into  the  solution,  cover 
immediately  with  a  glass  plate,  wrap  the  whole  in  a  towel,  and  set 
where  it  will  not  be  disturbed  until  the  next  exercise.  Remove  the 
few  large  crystals  formed  and  preserve  them.  Evaporate  the 
mother  liquor  not  quite  to  J  its  bulk  (say  T\),  and  set  this  to  crys- 
tallize in  exactly  the  same  manner  as  before.  Add  the  crystals 
so  obtained  to  the  first  lot.  Dry  the  crystals  on  paper  towels  and 
preserve  them  in  an  8-ounce  cork-stoppered  bottle. 

QUESTIONS 

1.  Of  what  does  the  insoluble  residue  consist  which  remains 
after  boiling  the  cryolite  with  milk  of  lime?  When  this  is  dis- 
carded after  partial  washing  according  to  directions,  what  pro- 


ALUM  AND  AMMONIA  FROM   ALUMINUM   NITRIDE      185 

portion  of  the  soluble  aluminum  salt  is  lost  with  it?  (Note  that 
the  actual  solid  material  of  the  slime  in  question  occupies  an 
inappreciable  volume  as  compared  with  the  liquid  in  which  it  is 
suspended,  even  after  the  slime  has  been  settling  for  several  days. 
See  Note  5  (6)  on  page  10.) 

2.  State  what  ions  are  produced  by  aluminum  hydroxide  (1) 
when  it  acts  as  an  acid;  (2)  when  it  acts  as  a  base.  Compare  its 
strength  as  an  acid  and  as  a  base  with  that  of  other  common 
electrolytes. 

Define  an  amphoteric  substance. 

PREPARATION  25 
ALUM  AND  AMMONIA  FROM  ALUMINUM  NITRIDE 

The  preparation  of  aluminum  nitride,  No.  13,  illustrates  one  of 
the  possibilities  in  the  way  of  fixation  of  atmospheric  nitrogen. 
Although  this  method  would  hardly  be  of  economic  importance 
unless  metallic  aluminum  could  be  produced  very  much  more 
cheaply,  it  is  obvious  that  if  the  aluminum  nitride  were  to  be 
utilized  as  a  source  of  ammonia  it  would  be  profitable  to  recover  a 
useful  compound  of  aluminum.  Boiling  with  a  solution  of  caustic 
alkali,  and  still  better,  fusion  with  the  dry  reagent  attacks  the 
nitride. 

A1N  +  3KOH  -»  KsAlOs  +  NH3. 


For  use  as  fertilizer  ammonia  is  invariably  absorbed  in  sulphuric 
acid,  and  put  on  the  market  as  the  solid  and  easily  handled  am- 
monium sulphate. 

Treatment  of  KsAlOs  with  sufficient  sulphuric  acid  yields  a 
solution  containing  the  constituents  of  potash  alum,  K2S04.A12- 
(804)3,  24H2O,  but  with  three  times  the  necessary  proportion  of 
K2S04.  Nevertheless,  pure  alum  can  be  crystallized  from  this 
solution.  The  mother  liquor  containing  the  excess  of  potassium 
sulphate  is  rejected.  It  is  impossible  to  avoid  this  excess  of 
potassium  sulphate  because  the  amount  of  caustic  alkali  prescribed 
is  the  minimum  which  will  give  a  satisfactory  reaction  with  the 
aluminum  nitride. 

Materials:     aluminum  nitride,  A1N,  10  grams  =  \  F.W.  (prod- 

uct of  Preparation  13) 

potassium  hydroxide,  in  sticks,  42  grams  =  J  F.W. 
6-normal  H2SO4,  250  cc. 


186  ELEMENTS  OF  GROUP   III 

Apparatus:  400  cc.  sheet  iron  crucible. 
5-inch  funnel. 

strip  of  asbestos  paper  for  gasket. 
2  filter  bottles  to  use  as  absorption  flasks. 
2  1-hole  rubber,jstoppers  to  fit  filter  bottles, 
connections  as  'described. 
600  cc.  beaker. 

1-liter  flask.  ^ 

8-inch  crystallizing  dish. 

Procedure:  Place  the  aluminum  nitride  and  the  potassium  hy- 
droxide in  the  iron  crucible.  Wet  the  strip  of  asbestos  paper  and 
fold  it  over  the  edge  of  the  crucible  all  the  way  around.  Press 
the  funnel  down  over  this  asbestos  gasket  to  make  a  tight  joint 
and  connect  the  stem  of  the  funnel  with  the  absorption  flasks, 
in  the  first  of  which  is  40  cc.  of  water,  and  in  the  second  15  cc.  of 
6-normal  H2SO4.  The  gas  delivery  tube  should  come  within  one 
half  inch  of  the  surface  of  the  liquid  in  each  bottle,  but  should 
not  dip  in  the  liquid.  The  delivery  tubes  should  be  of  good  size 
so  that  no  gas  pressure  can  arise  to  force  the  ammonia  out  past 
the  gasket.  Heat  the  crucible  with  a  very  low  flame  for  20 
minutes,  then  increase  the  size  of  the  flame  gradually  and  heat 
the  bottom  of  the  crucible  to  redness  for  30  minutes.  Pour  the 
contents  of  the  two  flasks  together  and  add  sulphuric  acid,  noting 
amount,  until  the  exact  point  of  neutrality  is  reached  (Note  11, 
page  17).  Evaporate  the  solution  on  steam  bath  to  complete 
dryness  (Note  6(6),  page  12),  and  put  up  the  ammonium  sulphate 
in  a  cork  stoppered  test  tube. 

After  the  crucible  has  cooled  add  250  cc.  of  water  and  let  the 
contents  disintegrate.  Pour  the  solution  and  sludge  into  a 
beaker  and  with  50  cc.  of  water  rinse  the  crucible  into  the  beaker. 
Boil  the  contents  of  the  beaker  5  minutes  and  filter  the  hot  solu- 
tion into  a  liter  flask.  Cool  the  solution  completely.  Hold  about 
J  of  it  in  reserve  and  add  6-normal  H2SO4  slowly  to  the  f ,  rotating 
the  flask  in  a  pan  of  cold  water  to  keep  the  solution  from  growing 
hot.  Record  in  the  note  book  the  volume  of  acid  which  first 
gives  a  precipitate  that  will  not  redissolve  on  rotating  the  contents 
of  the  flask  persistently,  and  then  that  will  just  redissolve  the 
precipitate.  Then  add  the  remaining  J  of  the  solution,  which 
will  bring  down  a  heavy  precipitate  again;  and  add  H2S04  very 


HYDRATED  ALUMINUM  CHLORIDE  187 

carefully  with  constant  rotation  until  this  precipitate  just  re- 
dissolves.  Record  the  total  volume  of  acid  used.  Filter  the 
solution  if  it  is  not  perfectly  clear  and  set  it  in  an  uncovered  8-inch 
crystallizing  dish,  in  a  place  free  from  dust,  to  evaporate  slowly 
and  crystallize.  If  preferred  the  method  of  crystallizing  outlined 
in  the  preceding  preparation  may  be  followed  (Note  8,  page  13, 
on  crystallization).  Discard  the  last  360  cc.  of  the  mother  liquor 
since  this  will  contain  so  much  excess  of  potassium  sulphate  that 
this  salt  would  crystallize  also.  Dry  the  alum  crystals  on  paper 
towels  and  preserve  them  in  an  8-ounce  cork-stoppered  bottle. 

QUESTIONS 

1.  The  aluminum  nitride  prepared  by  method  of  No.  13  con- 
tains aluminum  carbide  as  impurity.     Find  in  a  reference  book 
the  formula  of  this  carbide.     What  is  the  product  of  the  hydrol- 
ysis of  the  carbide?     Equation? 

2.  Answer  question  2  under  No.  24. 

3.  Write  in  ionic  form  the  successive  reactions  as  sulphuric  is 
gradually  added  to  potassium  aluminate  solution.     Explain  each 
reaction  and  state  to  what  type  it  belongs. 

4.  Repeat  question  3  for  the  gradual  addition  of  potassium 
hydroxide  to  an  aluminum  sulphate  solution. 

PREPARATION  26 

HYDRATED  ALUMINUM  CHLORIDE,  A1C13.6H2O 
A  solution  of  aluminum  chloride  can  be  prepared  by  the  action 
of  hydrochloric  acid  on  the  metal,  but  if  this  solution  is  evaporated 
to  dry  ness,  the  solid  that  is  left  is  the  oxide  instead  of  the  chloride. 
Hydrolysis  is  prevented  by  hydrochloric  acid,  and  the  hydrated 
chloride  can  be  crystallized  from  an  acid  solution.  In  this  prepa- 
ration the  solution  is  saturated  with  hydrochloric  acid,  which  not 
only  drives  back  hydrolysis  but  also  reduces  the  solubility  of  the 
salt. 

Anhydrous  aluminum  chloride  can  be  prepared  by  the  action 
of  dry  chlorine  on  aluminum;  in  its  properties  it  is  very  similar  to 
aluminum  bromide  which  is  the  subject  of  the  next  preparation. 
Materials:     aluminum  turnings,  13J  grams  =  ^  F.W. 
12-normal  HC1,  125  cc. 

90  grams  rock  salt  and  175  cc.  36-n  H2S04  for  hy- 
drogen chloride  generator. 


188  ELEMENTS  OF  GROUP  III 

shredded  asbestos. 

potassium  hydroxide  in  sticks  for  dessicator. 

ice. 
Apparatus:  750  cc.  flask. 

suction  filter  an<J  marble. 

1500  cc.  flask  for  generator  (Fig.  15  on  p.  118). 

3-inch  dish. 

dessicator  (a  wide  jar  covered  with  a  sheet  of 
asbestos  paper  pressed  close  to  the  edge  with  a 
heavy  plate  will  serve). 

Procedure:  Place  the  aluminum  turnings  in  a  750  cc.  flask,  add 
50  cc.  of  water  and  then  the  12-n  HC1  drop  by  drop  until  a  vigorous 
reaction  has  started  and  finally  as  rapidly  as  may  be  without  pro- 
ducing too  violent  a  reaction.  The  125  cc.  of  acid  should  just 
suffice  to  dissolve  the  metal.  Unless  it  is  perfectly  clear  filter  the 
solution  through  asbestos  on  a  suction  filter  (Note  4 (a),  page  8) 
and  return  it  to  the  750  cc.  flask.  Fit  up  the  hydrogen  chloride 
generator  and  connections  so  that  the  gas  will  pass  first  through 
a  washing  bottle  containing  a  little  12-n  HC1  and  provided  with  a 
safety  tube  (Fig.  19  on  p.  164).  The  gas  will  then  pass  into  the 
flask  containing  the  aluminum  chloride  solution.  The  end  of  the 
delivery  tube  dipping  into  the  solution  must  be  at  least  1J  cm.  in 
diameter,  else  it  will  become  stopped  with  the  precipitated  prod- 
uct. To  the  latter  flask  fit  an  exit  tube  which  will  lead  any 
waste  gas  to  within  J  inch  of  the  surface  of  water  in  a  bottle. 

Surround  the  flask  of  aluminum  chloride  with  cracked  ice  and 
water  and  pass  the  gas  into  the  solution  until  it  is  saturated. 
Observe  the  usual  caution  about  disposing  of  the  hot  contents  of 
the  generator.  Collect  the  crystalline  precipitate  in  a  5-inch 
funnel  containing  a  marble  and  dry  it  as  completely  as  possible 
with  suction  while  pressing  the  crystal  mass  with  the  round  end  of 
a  test  tube.  Place  the  product  in  a  3-inch  dish  in  the  dessicator 
in  which  sticks  of  potassium  hydroxide  have  been  placed.  It  will 
take  several  days  for  the  excess  of  hydrochloric  acid  to  evaporate 
from  the  crystals.  When  it  is  dry  put  the  product  in  a  6-ounce 
cork-stoppered  bottle. 

QUESTIONS 

1.  Treat  some  of  the  aluminum  chloride  with  water.  Does  it 
dissolve  to  give  a  clear  solution?  Is  there  any  noticeable  heating? 
Test  the  solution  with  litmus.  Is  the  salt  hydrolyzed?  Is  it 


ANHYDROUS  ALUMINUM  BROMIDE  189 

hydrolyzed  extensively?     Explain  how  you  reach  your  conclusion 
from  this  experiment. 

2.  Warm  about  1  gram  of  the  preparation  in  a  casserole  over  a 
flame  until  fumes  cease  to  come  off.     What  are  the  vapors  given 
off  and  by  what  tests  do  you  make  your  conclusion?     Is  the 
residue  soluble  in  water?     Of  what  does  it  consist? 

3.  Explain  the  difference  in  the  extent  to  which  the  hydrolysis 
takes  place  when  the  salt  is  dissolved  in  a  large  amount  of  water 
and  when  it  is  heated  with  merely  its  water  of  crystallization. 

PREPARATION  27 
ANHYDROUS   ALUMINUM   BROMIDE 

Like  aluminum  chloride  this  salt  is  completely  hydrolyzed  if  its 
solution  is  evaporated  to  dryness.  Only  the  hydrated  salt  can  be 
prepared  by  the  action  of  hydrobromic  acid  solution  (see  preceding 
preparation);  the  anhydrous  substance  is  prepared  by  the  direct 
action  of  the  elements  on  each  other.  The  action  is  one  of  such 
extreme  violence  that  it  is  safe  to  let  the  vapor  only  of  bromine 
come  in  contact  with  the  metal.  The  fact  that  aluminum  bromide 
can  be  purified  by  distillation  is  rather  typical  of  the  binary 
halogen  compounds,  most  of  which  are  volatile,  —  usually  very 
much  more  so  than  the  oxides. 

Aluminum  bromide  melts  at  90°  and  boils  at  265°.  It  is  very 
soluble  in  carbon  disulphide  without  chemical  action.  It  reacts 
explosively  with  water,  and  throughout  this  preparation  extreme 
caution  must  be  observed  not  to  let  any  of  it  come  in  contact  with 
water. 

Materials:     aluminum  turnings,  9  grams  =  ^  F.W. 

bromine,  80  grams  =  25  cc.  (must  be  dry  and  must 

be  measured  in  a  dry  graduate). 
Apparatus:  2  100  cc.  distilling  flasks. 

100  cc.  dropping  funnel. 

600  cc.  beaker. 

3  clean  dry  test  tubes. 

18  inches  hard  glass  tube  of  f-inch  diameter. 

Procedure:  This  preparation  must  be  carried  out  entirely  under 
the  hood.  Fit  the  dropping  funnel  with  a  rubber  stopper  into 
one  of  the  distilling  flasks,  clamp  the  latter  upright  with  the  bulb 


190  ELEMENTS  OF  GROUP  III 

immersed  in  water  boiling  in  a  600  cc.  beaker  on  a  lamp  stand. 
Fit  the  side  arm  of  the  distilling  flask  with  a  rubber  stopper  into 
one  end  of  the  hard  glass  tube.  The  other  end  of  this  tube  should 
be  thrust  loosely  into  the  neck  of  the  other  distilling  flask;  if  it 
is  too  large  it  must  be  .drawn  ojit  a  little  smaller  in  the  blast  flame. 
The  hard  glass  tube  should  -slope  somewhat  from  the  distilling 
flask  to  the  receiving  flask,  but  not  enough  so  that  the  aluminum 
turnings,  which  are  to  be  placed  about  in  th.e  middle  of  this  tube 
but  a  little  nearer  the  upper  end,  will  not  stay  in  place. 

The  entire  apparatus  must  be  absolutely  dry  inside.  Place  the 
bromine  in  the  dropping  funnel,  and  open  the  stop  cock  so  that  it 
drips  very  slowly  into  the  heated  flask  and  vaporizes.  Heat  the 
aluminum  fairly  strongly  in  the  hard  glass  tube  until  it  begins  to 
glow;  then  regulate  the  dripping  of  the  bromine  to  keep  up  a 
vigorous  reaction.  The  reaction  will  now  be  self  sustaining  if 
the  bromine  is  perfectly  dry.  If  it  is  necessary  in  order  to  sustain 
the  reaction,  the  glass  tube  may  be  heated.  Some  uncombined 
bromine  will  pass  the  aluminum  and  escape  from  the  receiving 
flask.  The  aluminum  bromide  will  condense  and  drip  into  the 
receiving  flask.  If,  as  is  not  likely,  any  solidifies  in  the  tube  it 
should  be  melted  out. 

After  this  phase  of  the  process  is  over  clamp  the  receiving  flask 
in  an  upright  position,  using  it  now  as  a  distilling  flask,  and  close 
the  neck  with  a  rubber  stopper.  Label  three  dry  test  tubes  and 
record  the  weight  of  each  on  the  label.  Support  one  of  these  tubes 
with  the  side  arm  of  the  distilling  flask  inserted  about  three  inches, 
and  start  the  distillation.  At  first  bromine  vapor  alone  comes 
over,  part  of  which  condenses.  As  soon  as  the  red  color  has  dis- 
appeared from  the  bulb  of  the  flask,  substitute  a  fresh  tube  as 
receiver,  and  catch  one  or  two  cc.  of  slightly  colored  product;  then, 
as  soon  as  the  aluminum  bromide  comes  over  entirely  colorless, 
change  the  receiver  again,  and  collect  the  bulk  of  the  product  in 
the  third  tube.  Stopper  all  three  tubes  immediately  with  rubber 
stoppers.  Hand  in  the  third  tube  and  contents  as  the  preparation. 
Half  fill  the  second  tube  with  carbon  disulphide  and  restopper  it, 
leaving  the  aluminum  bromide  to  dissolve.  Take  the  stopper 
from  the  distilling  flask  and  observe  the  dense  fumes,  caused  by 
the  reaction  of  the  aluminum  bromide  left  in  it  with  the  water 
vapor  of  the  air.  It  is  unsafe  to  pour  water  into  the  vessels  con- 
taining any  aluminum  bromide.  To  clean  them  dissolve  the 


EXPERIMENTS  191 

aluminum  bromide  in  a  little  carbon  disulphide,  then  add  water 
and  rinse  them  out.  If  a  deposit  of  aluminum  oxide  adheres 
inside  the  distilling  flask  add  a  few  cc.  of  12-n  HC1  and  let  it  stand 
over  night. 

QUESTIONS 

1.  Place  3  cc.  of  the  carbon  disulphide  solution  of  aluminum 
bromide  in  a  dry  600  cc.  beaker.     In  another  beaker  place  100  cc. 
of  cold  water  and  holding  it  at  arms  length  with  the  face  turned 
away,  pour  it  all  at  once  into  the  first  beaker.     The  reaction  is 
startling  but  not  dangerous  if  one  is  not  too  near.     Heat  the  con- 
tents of  the  beaker  until  the  carbon  disulphide  is  all  evaporated. 
Is  the  aluminum  bromide  now  dissolved  in  the  water?     Make 
tests  for  Al+++  and  Br~  ions  on  separate  small  portions.     How? 
Is  the  solution  clear,  or  cloudy?     Is  the  salt  in  the  dilute  aqueous 
solution  extensively  hydrolyzed? 

2.  Place  another  3  cc.  of  the  carbon  disulphide  solution  in  a 
watch  glass.     What  causes  the  dense  fogging?     After  the  carbon 
disulphide  evaporates  let  the  residue  stand  on  a  hot  plate  until 
it  appears  dry  and  no  longer  gives  off  a  fog.     Is  this  residue  soluble 
in  water?     Of  what  does  it  consist? 

3.  How  could  hydrated  aluminum  bromide  be  prepared? 

4.  Suggest  a  method  of  making  anhydrous  aluminum  bromide, 
using  hydrogen  bromide  instead  of  bromine,  and  state  why  you 
think  the  method  might  be  feasible. 

Experiments 

1.  Acid  Strength  of  Boric  Acid.  Dissolve  3  grams  of 
H3BO3  in  50  cc.  of  water,  thus  making  a  formal  solution. 
Add  a  few  drops  of  a  solution  of  blue  litmus  and  compare  the 
color  with  that  produced  by  dilute  HC1.  Add  normal  NaOH, 
one  cc.  at  a  time,  noting  the  gradual  change  in  color,  until 
the  litmus  is  completely  blue,  and  note  the  amount  taken. 

It  would  take  150  cc.  of  normal  NaOH  to  give  Na3BO3,  50  cc. 
to  give  NaB02,  and  25  cc.  to  give  Na2B4O7.  The  fact  that  the 
solution  becomes  alkaline  when  there  is  still  a  large  excess  of  boric 
acid  in  the  solution  (even  when  it  is  figured  as  tetraboric  acid, 
H2B407)  shows  how  very  weak  the  acid  is. 


192  ELEMENTS  OF  GROUP  III 

AMPHOTERIC  SUBSTANCES 

A  substance  which  can  behave  both  as  an  acid  and  a  base  is 
known  as  amphoteric.  In  water  such  a  substance  would  yield 
both  H+  and  OH~  ions,  but  the  product  of  the  concentrations  of 
these  ions  could  not  exceed  thQion  product  of  water; 

(H+)  X  (OH-)  =  10-7  X  10-7  =  10-1 

Amphoteric  substances  are  necessarily  extremely  weak  both  as 
acids  and  as  bases,  but  they  possess  the  property  of  reacting  with 
strong  acids  on  the  one  hand  and  with  strong  bases  on  the  other 
hand. 

Aluminum  hydroxide  is  amphoteric;  it  is  itself  insoluble,  but 
its  salts,  such  as  A1C13  and  Na3A103,  are  very  soluble;  it  is  therefore 
most  easy  to  observe  whether  the  aluminum  hydroxide  reacts  with 
an  acid  or  a  base,  because  if  it  does  it  is  seen  to  dissolve. 

2.  Aluminum  Hydroxide.    Add  NH4OH  to  an  aluminum 
salt  solution  (nitrate,  chloride,  or  sulphate)  until  a  strong 
odor  of  ammonia  persists  after  stirring.     A  gelatinous  white 
precipitate  forms.     Collect   some   of  this   precipitate   on  a 
filter  and  wash  it  repeatedly  with  hot  water  until  all  excess 
of  ammonia  is  removed.     Then  dip  red  and  blue  litmus  paper 
into  the  precipitate  and  observe  that  neither  is  affected. 

This  experiment  shows  the  insolubility  of  aluminum  hydroxide 
and  the  extreme  weakness  of  its  acid  and  basic  properties. 

3.  Aluminum  Hydroxide  Acting  as  an  Acid.    To  a  little 
aluminum  salt  solution  (nitrate,  chloride,  or  sulphate)  add 
NaOH  solution  a  drop  at  a  time  until  a  copious  precipitate 
is  observed;    continue  to  add  NaOH  and  observe  that  the 
precipitate  soon  redissolves. 

The  precipitate  is  aluminum  hydroxide,  A1(NO3)3  +  3NaOH  — » 
A1(OH)3  +  3NaNO3.  That  the  aluminum  hydroxide  thus  dis- 
solves in  the  solution  of  the  base,  indicates  that  it  has  reacted  to 
form  a  soluble  salt  and  that  it  is  itself  an  acid.  There  are  two 
acids  of  aluminum  differing  in  degree  of  hydration  in  the  same 
way  as  the  different  boric  acids. 

Al(OH),  -»  H3A103  (aluminic  acid). 

A1(OH)3  ->  H20  +  HA!02  (meta  aluminic  acid). 


EXPERIMENTS  193 

The  soluble  salt  may  be  either  sodium  ortho  aluminate,  Na3AlO3, 
or  sodium  meta  aluminate,  NaAlO2,  depending  on  the  amount  of 
NaOH  used. 

4.  Aluminum  Hydroxide  Acting  as  a  Base.     To  the  solu- 
tion left  at  the  end  of  Exp.  3  add  an  acid  (say  HNO3)  drop 
by  drop  until  a  copious  precipitate  is  observed.     Continue 
to  add  acid  and  observe  that  the  precipitate  redissolves. 

The  precipitate  is  aluminic  acid,  displaced  from  its  salt  by  the 
stronger  acid,  Na3AlO3  +  3HNO3  ->  H3A1O3  +  3NaNO3.  That 
the  aluminic  acid  dissolves  in  nitric  acid  indicates  that  it  reacts  to 
form  a  soluble  salt  and  is  itself  reacting  as  a  base. 

5.  Comparative  Basic  and  Acidic   Strength  of  A1(OH)3. 
Add  NH4OH  in  excess  to  an  aluminum  salt  solution,  and  note 
that  the  precipitate   does  not  redissolve   in  the   excess  of 
NH4OH.     Add  acetic  acid  drop  by  drop.     For  a  time  the 
precipitate  remains  unaffected,  but  after  the  NH4OH  is  all 
neutralized  (shown  by  the  solution  ceasing  to  smell  of  am- 
monia after  shaking)  a  little  more  acetic  acid  redissolves  the 
precipitate. 

Ammonium  hydroxide  and  acetic  acid  are  of  equal  strength  as 
base  and  acid  respectively.  That  the  A1(OH)3  does  not  dissolve 
in  excess  NH4OH  indicates  that  it  is  not  a  strong  enough  acid  to 
react  with  a  base  of  this  feeble  strength.  That  it  does  dissolve 
in  acetic  acid  indicates  that  it  is  a  strong  enough  base  to  react  with 
an  acid  of  the  feeble  strength  of  acetic  acid.  Thus  aluminum 
hydroxide,  although  an  extremely  weak  base,  is  more  basic  than 
acidic. 

6.  Instability    of    Aluminum    Carbonate.     Add    together 
solutions    of   A1(N03)3    and    Na^COs.     Effervescence    takes 
place  and  the  escaping  gas  can  be  shown  to  be  carbon  dioxide. 
A  gelatinous  white  precipitate  is  formed  resembling  previous 
precipitates  of  A1(OH)3.     To  show  whether  this  precipitate 
is  really  the  hydroxide,  or  perhaps  the  carbonate  or  basic 
carbonate,  collect  some  of  it  on  a  filter  and  wash  it  very 
thoroughly  with  water  to  remove  any  excess  of  Na2CO3. 
Then  add  an  acid  (say  HC1)  to  the  precipitate  and  observe 
that  it  dissolves  without  a  trace  of  effervescence. 


194  ELEMENTS  OF  GROUP  III 

Reference  books  tell  us  that  aluminum  carbonate  has  never 
been  prepared.  In  this  experiment  we  brought  together  the  ions 
of  aluminum  carbonate  and  the  result  showed  that  this  salt  cannot 
exist  in  solution.  The  reaction  might  be  described  as  the  hydrol- 
ysis of  aluminum  carbonate : 

2A1+++  3CO8— 

6H2O-»6OH-  6H  + 

I  I 

A1(OH)8 1       3H2CO3 

I 
3H2O  +  3CO2 1 

That  aluminum  carbonate  is  thus  completely  hydrolyzed  indicates 
that  the  base  and  the  acid  are  both  very  weak. 

GENERAL  QUESTIONS  VI 
ELEMENTS  OF  GROUP  III 

1.  Make  a  table  of  all  the  elements  of  Group  III,  giving  in 
succeeding  columns:    1,  the  symbols  of  the  elements  in  the  order 
of  their  increasing  atomic  weights,  placing  the  elements  of  the  A 
Family  in  the  left  of  the  column,  and  of  the  B  Family  in  the  right; 
2,  the  formula  of  the  hydroxide,  or  hydroxides,  if  more  than  one 
is  described  in  reference  books;  3,  the  character  of  the  hydroxide, 
distinguishing,  strongly  acid,  weakly  acid,   amphoteric,   weakly 
basic,  and  strongly  basic;  4,  the  formula  of  the  chloride,  or  chlor- 
ides if  more  than  one  is  described  in  the  reference  books;   5,  the 
degree  of  hydrolysis  of  the  chloride,   distinguishing,   complete, 
much,  little,  or  none. 

2.  Compare  aluminum  hydroxide  with  the  hydroxides  of  sodium 
and  magnesium,  the  corresponding  elements  in  Groups  I  and  II, 
as  regards  solubility  and  degree  of  ionization. 

3.  Compare  the  stability  of  magnesium  and  aluminum  car- 
bonates, —  of  dry  magnesium"  -and  aluminum  sulphates.     Find 
the  heat  of  formation  of  equivalent  amounts  of  the  sulphates 
according  to  the  equations. 

3MgO  +  3SO3  ->  3MgSO4. 
A1203  +  3SO3  -*  A12(S04)3. 


GENERAL  QUESTIONS  VI  195 

What  do  the  facts  thus  cited  show  as  to  the  relative  basic  strength 
of  magnesium  and  aluminum  oxides.  (Compare  General  Ques- 
tions V,  Nos.  5  and  6.) 

4.  Describe  the  successive  observations  that  are  made  when 
NaOH  solution  is  added  gradually  to  an  A1C13  solution  until  an 
excess  of  the  base  is  present.  Likewise  describe  observations 
when  HC1  solution  is  added  gradually  to  an  NasAlOs  solution 
until  the  acid  is  in  excess.  Write  fully  ionized  intersecting 
equations  corresponding  to  each  observation,  and  explain  how  the 
amphoteric  character  of  aluminum  hydroxide  is  manifested  in  this 
experiment. 


CHAPTER  VII 

HEAVY  METALS  OF  GROUPS  I  AND  II  OF 
THE  PERIODIC   SYSTEM 

The  metals  coming  under  this  heading  constitute  the  right 
hand  or  B  Families  in  Groups  I  and  II  of  the  periodic  system. 
They  possess  high  specific  gravities,  and  chemically  they  are  far 
less  active  than  the  metals  of  the  corresponding  A  Families,  — 
they  being  not  greatly,  or  not  at  all,  affected  by  the  atmosphere 
or  by  water.  They  are  distinctly  base-forming  with  the  exception 
of  gold,  in  that  their  oxides  yield  fairly  stable  salts  with  the  strong 
acids;  but  their  basic  properties  are  comparatively  weak,  and  the 
oxides  of  some  of  them  show  very  feeble  acidic  properties  as  well. 

Copper,  silver,  and  gold  in  Group  I  show  a  similarity  to  sodium 
and  potassium  principally  in  the  fact  that  they  form  certain 
compounds  of  the  same  type,  for  example,  M20  and  MCI.  Zinc, 
cadmium,  and  mercury  in  Group  II  resemble  calcium,  barium,  and 
strontium  in  that  they  form  compounds  of  the  types,  MO,  MS04, 
MC12,  etc.  In  other  respects,  the  divergence  in  the  properties 
of  the  elements  of  the  A  and  B  Families  is  at  a  maximum  in  these 
two  groups. 

PREPARATION  28 

CRYSTALLIZED  COPPER  SULPHATE  (BLUE  VITRIOL) 
FROM  COPPER  TURNINGS 

On  account  of  the  fact  that  copper  has  not,  like  zinc,  iron,  etc., 
the  power  of  displacing  hydrogen  from  acids,  it  is  not  possible  to 
dissolve  it  directly  in  dilute  sulphuric  acid.  But  although  the 
metal  itself  is  so  difficult  to  affect  with  acids,  nevertheless  copper 
oxide  is  readily  dissolved;  and  thus  the  problem  is  to  convert 
copper  into  its  oxide.  The  cheapest  source  of  oxygen  is  the 
atmosphere,  and  on  the  commercial  scale,  the  usual  method  of 
obtaining  copper  sulphate  from  scrap  copper  is  to  allow  dilute 
sulphuric  acid  to  drip  slowly  over  the  latter,  to  which  air  is  given 
free  access.  Since,  however,  this  method  would-be  too  time- 

196 


COPPER  SULPHATE 


197 


consuming  for  the  laboratory,  nitric  acid  will  be  employed  instead 
of  air  as  the  oxidizing  agent. 

3Cu  +  2HNO3  =  3CuO  +  H:O  +  2NO. 
CuO  +  H2SO4  =  CuS04  +  H2O. 
Materials:     copper  turnings,  64  grams  =  1  F.W. 
6-normal  HNO3. 
6-normal  H2S04. 
Apparatus:  500  cc.  dish, 
suction  filter. 

6-inch  crystallizing  dish  with  watch  glass  or  glass 
plate  to  cover  it. 

Procedure:  Heat  the  copper  turnings  in  an  iron  pan  until  all 
oily  matter  is  burned  and  the  metal  has  become  coated  with  oxide, 
In  a  500  cc.  porcelain  dish  treat  the  ignited  copper  turnings  with 
the  calculated  volume  of  6-n  H2SO4  and  6-n  HNO3.  Warm  in  the 
hood  for  20  minutes;  if  any  metallic  copper  remains  undissolved, 
pour  the  solution  off  from  it  and  dissolve  it  with  a  few  cubic  centi- 
meters of  fresh  nitric  acid  and  twice  as  much  sulphuric  acid.  If 
the  solution  is  not  perfectly  clear,  filter  it  while  still  at  the  boiling 
temperature;  then  cool  as  rapidly  as  possible,  stirring  to  get  a 
crystal  meal.  Separate  the  meal  from  the  mother  liquor  using  the 
suction  filter.  Evaporate  the  mother  liquor  somewhat,  and  obtain 
a  second  crop  of  crystal  meal,  throwing  away  the  mother  liquor 
from  this  crystallization.  Dissolve  all  of  the  damp  product  by 
adding  its  own  weight  of  water  and  warming;  place  the  warm 
solution  in  the  crystallizing  dish,  add  some  seed  crystals  when  the 
temperature  is  about  35°,  cover  the  dish,  wrap  it  with  towels  and 
leave  it  to  cool  very  slowly.  Remove  and  save  the  crystals,  and 
leave  the  saturated  solution  in  the  crystallizing  dish  uncovered 
to  evaporate  slowly  and  deposit  more  crystals.  Spread  the  entire 
product  on  paper  towels  to  dry,  and  hand  in  'the  preparation  in  a 
12-ounce  cork  stoppered  bottle. 

QUESTIONS 

1.  Explain  why  copper  will  not  dissolve  in  dilute  sulphuric  acid. 

2.  Write  the  equation  for  the  reaction  of  copper  with  con- 
centrated sulphuric  acid.     Analyze  this  reaction  and  show  in  what 
manner  the  copper  is  oxidized. 

3.  How  can  copper  sulphate  be  obtained  from  copper  sulphide 
on  a  commercial  scale? 


198  HEAVY  METALS  OF  GROUPS  I  AND  II 

PREPARATION  29 
CUPROUS  CHLORIDE 

Although  copper  itself  is  not  readily  oxidized,  yet  when  oxida- 
tion is  once  induced  it  proceeds  under  most  conditions  at  once  to 
the  "  ic  "  state  (corresponding  to  the  oxide  GuO).  This  is  in 
accordance  with  the  fact  that  cuprous  salts  are  oxidized  with  the 
greatest  readiness,  so  that,  in  the  presence  of  any  oxidizing  agent 
powerful  enough  to  oxidize  metallic  copper,  any  cuprous  salt  which 
might  first  be  formed  would  instantly  be  oxidized  to  cupric. 
When  metallic  copper  is  placed  in  a  cupric  chloride  solution,  the 
former  soon  becomes  coated  with  a  white  deposit  of  cuprous 
chloride, 

Cu  +  CuCl2  -*  2CuCl  | 

the  copper  acting  as  a  reducing  agent,  and  the  cupric  ion  as  an 
oxidizing  agent.  This  insoluble  coating  soon  prevents  further 
contact  of  the  reacting  substances  with  each  other,  and  the  re- 
action becomes  very  slow,  or  ceases.  But  if  concentrated  HC1 
is  added  to  the  solution,  the  cuprous  chloride  is  dissolved  and  the 
reduction  of  the  cupric  salt  continues  to  completion.  The  cuprous 
chloride  and  chloride  ions  unite  to  form  the  complex  cupro- 
chloride  ions  CuCl2~  and  CuCl3  .  But  these  complex  ions  are 
not  very  stable,  and  if  the  concentration  of  chloride  ions  is  di- 
minished by  diluting  the  solution,  they  dissociate,  and  cuprous 
chloride  precipitates.  Cuprous  chloride  is  so  readily  oxidized 
that,  even  in  contact  with  moist  air  or  suspended  in  water  con- 
taining dissolved  air,  it  is  changed  to  a  cupric  salt: 
2CuCl  +  2HC1  +  O  =  H20  +  2CuCl2, 

or  2CuCl  +  H20  +  O  =  2Cu  <  ^  I 


Hence  a  great  deal  of  care  will  be  necessary  in  washing  and  drying 
the  cuprous  chloride  to  prevent  its  becoming  discolored  in  conse- 
quence of  oxidation.  When  perfectly  dry,  however,  or  when 
covered  with  dry  ether,  the  oxidation  takes  place  very  slowly. 

Materials:     cryst.  cupric  chloride,  CuCl2.2H2O,  43  g.  =  J  F.W. 
clean  copper  wire,  25  grams. 
12-normal  HC1  100  cc. 
95%  alcohol,  30  cc. 
ether,  30  cc. 


CUPROUS  CHLORIDE  199 

Apparatus:  300  cc.  flask. 

short  stemmed  funnel. 
2-liter  common  bottle, 
suction  filter. 

Procedure:  Dissolve  the  crystallized  cupric  chloride  in  100  cc.  of 
water  and  100  cc.  of  12-n  HC1.  Place  the  solution,  together  with 
the  clean  copper  wire  in  a  flask  and  suspend  a  short-stemmed 
funnel  in  the  neck  of  the  flask  to  prevent  in  part  the  escape  of  acid 
vapors.  Warm  the  mixture  in  the  flask  to  about  80°  and  hold 
it  at  that  temperature  by  placing  the  flask  over  a  very  tiny  flame 
or  on  a  warm  plate,  but  do  not  allow  the  solution  to  boil.  The 
green  color  disappears  gradually  and  the  solution  becomes  color- 
less, but  usually,  the  moment  this  point  is  reached,  an  intense 
brown  color,  arising  from  some  unexplained  impurity,  develops 
within  a  few  seconds,  and  unless  one  watches  all  the  time,  he  will 
fail  to  see  the  transition.  Test  for  complete  reduction  by  pouring 
J  cc.  of  the  solution  in  a  test  tube  and  adding  water.  White 
cuprous  chloride  is  precipitated  and  if  the  cupric  salt  is  all  reduced 
the  solution  will  show  no  blue  color.  Pour  the  contents  of  the 
tube  into  the  2-liter  bottle,  and  if  the  test  was  satisfactory,  put 
2  liters  of  cold  water  in  the  bottle  and  pour  into  it,  in  a  thin  stream, 
all  the  solution  from  the  flask,  leaving  the  remnants  of  copper  wire 
behind.  Rinse  the  flask  into  the  bottle  with  about  25  cc.  of  water. 
Mix  the  contents;  let  the  white  precipitate  settle;  pour  off  most 
of  the  liquid,  stir  up  the  precipitate  with  the  rest  and  pour  the 
suspension  into  the  suction  filter;  add  a  few  drops  of  HC1  and 
50  cc.  of  water  to  the  bottle  and  rinse  out  the  last  of  the  cuprous 
chloride  into  the  filter.  Continue  the  suction  until  the  last  drop 
of  liquid  is  drawn  into  the  cake  of  cuprous  chloride;  stop  the  suc- 
tion; pour  15  cc.  of  alcohol  in  a  thin  stream  around  the  upper 
edge  of  the  funnel  and  let  it  run  down  washing  the  sides  of  the 
funnel  and  covering  the  cuprous  chloride.  Apply  the  suction 
until  the  last  drop  of  alcohol  is  sucked  into  the  cake,  then  stop  the 
suction.  In  exactly  the  same  way,  wash  the  sides  of  the  funnel 
and  the  cuprous  chloride  with  one  more  15  cc.  portion  of  alcohol 
and  with  two  successive  15  cc.  portions  of  ether.  Transfer  the 
caked  cuprous  chloride  to  paper  towels,  break  up  the  cake  with 
a  spatula,  and  leave  the  preparation  on  the  warm  plate  5-10 
minutes,  until  the  adhering  ether  has  evaporated.  Pulverize 


200  HEAVY  METALS  OF  GROUPS  I  AND   II 

the  product  and  put  it  up  in  a  warmed  (to  thoroughly  dry) 
6-inch  cork  stoppered  test  tube.  The  product  should  be  pure 
white. 

QUESTIONS 

1.  Metallic  copper  placed  ;in  a  cupric  sulphate  solution  causes 
no  easily  observed  reaction,  as  is  probably  well  known  to  the 
student.     Nevertheless,    very    careful    experiments    are    capable 
of  showing  that  cuprous  ions,  Cu+,  are  formed  to  a  very  small 
extent  in  the  solution,  Cu°  +  Cu++  +±  2Cu+.     But  this  reaction 
is  reversible  and  it  has  already  reached  a  state  of  equilibrium 
when  the  concentration  of  cuprous  ions  is  so  small  as  to  escape 
detection  except  by  the  most  refined  methods  of  measurement. 

In  view  of  the  fact  that  cuprous  chloride  is  extremely  insoluble, 
state  why  the  above  reaction  can  proceed  to  completion  towards 
the  right  in  a  solution  containing  chloride  ions. 

2.  Place  2  to  3  grams  of  cuprous  chloride  in  the  bottom  of  a  dry 
test  tube.     Fill  it  completely  with  6-n  NH4OH  and  immediately 
stopper  it  tightly,  allowing  no  air  bubble  to  remain  at  the  top. 
Invert  the  tube  a  number  of  times  until  the  salt  is  dissolved.     At 
this  point,  the  solution  should  be  nearly  colorless,  and  it  would 
be  quite  so  if  the  salt  had  been  pure  and  air  had  been  completely 
excluded.     Unstopper  the  tube  and  quickly  pour  its  contents 
into  a  bottle  of  about  300  cc.  capacity,  and  immediately  cork  the 
latter  air-tight.     Shake  the  bottle  vigorously  for  3  minutes,  then 
place  the  mouth  under  water  and  remove  the  cork.     From  its 
quantity,  infer  what  gas  is  absorbed  from  the  air  in  the  bottle. 
What  is  the  change  in  the  condition  of  the  copper  salt  which  causes 
its  change  in  color?     Equations? 

3.  Spread  about  J  gram  of  cuprous  chloride  in  a  watch  glass; 
moisten  it,   and  let  it   stand   5-10  minutes.     What  causes  the 
discoloration?     Equation?     Rinse    the    discolored    mass    into    a 
beaker  and   add    1-2   cc.  6-n  HC1.     What  causes  the  solid  to 
again  become  white  and  the  solution  blue?    Equation? 

PREPARATION  30 
CUPROUS  OXIDE 

When  metallic  copper  is  heated  in  the  air  it  becomes  coated 
with  a  layer  of  oxide,  which,  according  to  conditions,  may  be 
cuprous  or  cupric  oxide,  or  a  mixture  of  the  two.  Pure  cuprous 


CUPROUS  OXIDE  201 

oxide  is  most  conveniently  prepared  in  the  wet  way  by  treating 
an  alkaline  cupric  salt  solution  with  a  reducing  agent,  whereby 
the  red  cuprous  oxide  is  precipitated. 

Cupric  hydroxide  is  nearly  insoluble  in  alkalies  alone,  but  it 
dissolves  when  a  soluble  tartrate  is  added,  the  copper  going  into 
a  complex  negative  ion  similar  in  color  to  the  ammonio-cupric  ion. 
Such  a  solution  is  known  as  Fehling's  solution. 

Materials:     blue  vitriol,  CuSO4.5H2O,  50  grams  =  J  F.W. 

Rochelle  salt,  KNaC4H406,  75  grams. 

sodium  hydroxide  in  sticks,  75  grams. 

grape  sugar,  100  grams. 

alcohol,  45  cc. 
Apparatus:  8-inch  evaporating  dish. 

2-liter  common  bottle. 

suction  filter. 

350  cc.  casserole,  3  beakers,  and  3  8-inch  common 
filters. 

Procedure:  Dissolve  the  blue  vitriol  and  the  Rochelle  salt  each 
in  200  cc.  of  hot  water,  filter  the  solutions  separately  if  they  are 
not  clear,  and  when  they  are  nearly  cold  mix  them  together  in  the 
8-inch  dish.  In  the  meantime  dissolve  the  sodium  hydroxide  in 
200  cc.  of  water,  cool  it  completely,  filter  if  not  clear,  and  pour 
it  in  a  thin  stream  into  the  mixture  in  the  dish,  stirring  constantly 
and  being  careful  not  to  allow  the  mixture  to  get  more  than  barely 
warm  from  the  reaction.  A  clear  solution  results,  but  one  of  so 
deep  blue  a  color  as  to  appear  almost  like  ink.  Dissolve  the 
grape  sugar  in  150  cc.  of  water,  filter  if  necessary,  add  it  to  the 
solution  in  the  dish,  and  boil  the  mixture  until  the  copper  has  pre- 
cipitated as  the  bright  red  cuprous  oxide,  and  the  blue  color  has 
disappeared  from  the  liquid.  Transfer  the  contents  of  the  dish 
to  the  2-liter  bottle,  add  water  to  fill  the  latter,  mix  the  contents 
thoroughly,  and  let  settle  for  about  15  minutes.  Although  the 
supernatant  liquid  still  appears  turbid  at  this  point,  the  amount 
of  suspended  cuprous  oxide  is  very  small.  Decant  this  turbid 
liquid  very  carefully  so  as  not  to  stir  up  the  cuprous  oxide.  As 
the  bottle  is  gradually  tipped,  the  heavy  cuprous  oxide  finally 
settles  into  the  shoulder  of  the  bottle  and  all  but  about  100  cc. 
of  the  liquid  runs  over  the  lip.  Fill  the  bottle  again  with  water, 
mix,  let  settle,  and  decant  as  before.  Transfer  the  cuprous  oxide 


202  HEAVY  METALS  OF  GROUPS  I  AND  II 

to  the  suction  filter,  and  as  soon  as  the  last  drop  of  water  is  sucked 
into  the  compacted  cake  in  the  bottom,  stop  the  suction.  Then 
wash  with  2  successive  portions  of  15  cc.  each  of  water  and  3 
successive  portions  of  15  cc.  each  of  alcohol  following  the  directions 
of  the  preceding  preparation,  ji  Place  the  cuprous  oxide  on  a  folded 
paper  towel  on  the  warm  plate,  and  as  soon  as  the  alcohol  is  evap- 
orated, put  up  the  product  in  a  6-inch  cork  stoppered  test  tube. 

QUESTIONS 

1.  Treat  small  samples  of  the  product  with  6-n  H2S04,  6-n  HC1 
and  6-NH4OH  respectively.  What  reaction  would  you  expect  to 
take  place  with  these  reagents,  and  what  evidence  of  such  reac- 
tions do  you  actually  observe? 

Place  0.5  gram  of  cuprous  oxide  in  a  test  tube,  fill  the  tube  with 
6-n  HC1,  and  stopper  it  tightly  with  a  rubber  stopper.  Let  it 
stand  at  least  24  hours,  shaking  as  often  as  convenient,  and  letting 
the  tube  lie  on  its  side  the  rest  of  the  time  in  order  to  expose  a 
larger  surface  of  the  solid  material  to  the  reagent.  Finally  pour 
some  of  the  clear  solution  into  a  large  beaker  of  water.  Does  this 
furnish  any  evidence  that  the  cuprous  oxide  has  reacted  with  the 
acid? 

PREPARATION  31 
AMMONIO-COPPER  SULPHATE,  CuS04.4NH3.H20 

Preparation  3  illustrated  the  formation  of  a  double  salt,  potas- 
sium and  copper  sulphate,  and  one  of  the  experiments  following 
that  preparation  showed  that  ammonium  could  be  substituted  for 
potassium  to  give  the  ammonium  and  copper  sulphate,  (NH4)2S04. 
CuS04.6H2O.  In  the  double  salt,  ammonium  plays  the  part  of  a 
positive  radical.  In  the  present  preparation  ammonia  plays  an 
altogether  different  role.  It  does  not  possess  a  primary  valence, 
and  it  enters  into  a  molecular  compound  with  the  salt  in  virtue 
only  of  a  secondary  valence.  In  fact,  the  ammonia  in  this  prepa- 
ration is  held  in  the  same  sort  of  a  combination  as  the  water  in  the 
hydrate  CuS04.5H20.  The  molecules  of  ammonia  would  appear 
to  be  bound  to  the  copper  rather  than  to  the  sulphate  radical, 
because  when  the  salt  is  dissolved  in  water  the  four  ammonia 
molecules  remain  in  combination  with  the  copper  as  the  complex 
ion  Cu(NH3)4++,  while  the  sulphate  radical  appears  as  the  ordinary 


AMMONIO-COPPER  SULPHATE  203 

SO 4      ion.     Thus  we  might  say  that  this  salt  is  the  sulphate  of  the 
ammonio-copper  complex.     (Cf.  Ammoniates,  page  97.) 

The  salt  is  exceedingly  soluble  in  water,  and  therefore,  in  pre- 
paring it,  use  is  made  of  its  insolubility  in  alcohol.  The  method 
adopted  of  allowing  the  alcohol  to  mix  with  the  aqueous  solution 
by  slow  diffusion  is  to  insure  the  formation  of  large,  well-defined 
crystals. 

blue  vitriol,  CuSO4.5H2O,  25  grams  =  TV  F.W. 
6-normal  NH4OH,  85  cc. 
alcohol,  165  cc. 
ether,  20  cc. 
Apparatus:  100  cc.  dropping  funnel. 

300  cc.  tall  common  bottle. 
2  250  cc.  flasks. 
400  cc.  beaker. 

Procedure:  Pulverize  the  blue  vitriol,  place  it  in  a  flask  and  dis- 
solve it  in  the  85  cc.  of  cold  6-n  NH4OH.  (See  Note  7,  page  13.) 
Pour  the  solution  through  a  filter,  catching  the  filtrate  in  another 
flask.  Place  125  cc.  of  alcohol  in  the  bottle;  fill  the  stem  of  the 
dropping  funnel  with  water;  insert  it  to  the  bottom  of  the  alcohol, 
and  run  in  20  cc.  of  water  to  form  a  layer  beneath  the  alcohol  to 
separate  the  latter  from  the  ammoniacal  copper  solution,  which 
is  next  introduced  through  the  funnel.  Allow  no  bubbles  of  air 
to  be  sucked  with  the  liquid  into  the  stem  of  the  funnel  and  thus 
avoid  stirring  up  the  solution  and  mixing  the  layers.  Set  the 
bottle  away  for  at  least  a  week,  at  the  end  of  which  time  crystals 
2  and  3  cm.  long  will  have  formed.  The  alcoholic  and  aqueous 
layers  have  not  yet  completely  diffused  into  each  other,  and  when 
they  are  mixed  a  meal  of  very  small  crystals  is  precipitated. 
Therefore  pour  the  liquid  all  at  once  out  of  the  bottle  into  a  clean 
beaker.  The  large  crystals  adhere  to  the  inside  of  the  bottle. 
Remove  them  to  a  small  dish;  add  10  cc.  of  alcohol  to  which  1  cc. 
of  ammonia  has  been  added;  stir  thoroughly  by  rotating  the  dish, 
and  pour  off  the  alcohol,  allowing  it  to  carry  with  it  any  of  the 
precipitate  of  small  crystals.  Repeat  the  washing  with  10  cc.  of 
alcohol  and  then  with  10  cc.  of  ether.  Spread  the  crystals  on 
paper  towels  and  leave  them  until  they  cease  to  smell  of  ether. 
Then  at  once  stopper  them  tightly  in  a  6-inch  test  tube,  since 
they  give  off  their  ammonia  rather  easily.  Drain  on  a  suction 


204  HEAVY  METALS  OF  GROUPS  I  AND  II 

filter  the  crystal  meal  formed  in  the  beaker,  and  wash  it  on  the 
filter  (Note  5  (a))  with  the  same  liquids  as  were  used  for  the  larger 
crystals.  Preserve  the  large  crystals  and  the  crystal  meal  sepa- 
rately, each  in  a  6-inch  cork  stoppered  test  tube. 

QUESTIONS 

1.  What  is  the  reaction  between  Cu++  and  OH~  ions?    To 
a  little  l-nCuS04  add  6-n  NaOH  drop  by  drop,  until  it  is  present 
in  large  excess.     Save  the  solution. 

2.  When  ammonium  hydroxide  is  added   instead    of   sodium 
hydroxide,  explain  what  successive  reactions  occur. 

3.  Add  ammonium  chloride  to  the  solution  saved  from  (1)  and 
explain  the  results. 

4.  To  5  cc.  of  1-n  CuS04  add   10  cc.  of  a  molal  solution   of 
tartaric  acid;    then  add  sodium   hydroxide   solution,   as  in  (1), 
and  compare  the  results  with  those  in  (1)  and  (2),  but   do  not 
attempt  to  ascribe  a  definite  formula  to  the  complex  compound 
formed. 

PREPARATION  32 
ZINC  OXIDE 

Zinc  oxide  is  used  as  a  white  pigment,  for  which  purpose  it  has 
the  advantage  of  not  turning  black  under  the  action  of  hydrogen 
sulphide.  It  may  be  obtained  directly  by  burning  metallic  zinc, 
or  from  a  soluble  zinc  salt  by  precipitating  first  a  basic-  carbonate 
and  then  heating  this  to  convert  it  into  the  oxide.  Both  zinc 
carbonate  and  zinc  hydroxide  are  insoluble  in  water,  but  the  basic 
carbonate  is  of  still  greater  insolubility,  and  therefore  precipitates 
more  readily  than  either  of  the  former  when  the  ions  necessary 
to  its  formation  are  brought  together.  The  simplest  formula  for 
the  basic  carbonate  is 

OH 


Zn< 

7r> 

Zn 


QH 


but  the  precipitate  may  be  of  varying  composition  according  to  the 
conditions  of  its  formation. 

If  zinc  sulphate  in  solution  is  treated  with  sodium  bicarbonate, 
pure  zinc  carbonate  is  precipitated,  because  a  sodium  bicarbonate 
'solution  contains  but  a  minute  quantity  of  OH ~  ions.  On  the 


ZINC  OXIDE  205 

other  hand,  a  sodium  carbonate  solution,  in  consequence  of  hydrol- 
ysis, contains  a  considerable  quantity  of  OH~-ions,  and  thus  it 
furnishes  both  the  CO3 and  OH~-ions  necessary  for  the  forma- 
tion of  basic  zinc  carbonate. 

Zinc  carbonate  is  decomposed  by  heat  into  zinc  oxide  and  carbon 
dioxide. 

Commercial  zinc  sulphate  invariably  contains  a  small  amount  of 
iron  as  an  impurity.  Since  FeSO4.7H20  crystallizes  isomor- 
phously  with  ZnSO4.7H20  a  preparation  of  the  latter  cannot  be 
freed  of  the  former  by  recrystallization.  By  addition  of  chlorine 
water  to  the  solution  of  zinc  sulphate,  the  iron  is  oxidized  to  ferric 
salt;  the  ferric  salt  hydrolyzes  somewhat,  and,  if  the  acid  produced 
by  the  hydrolysis  is  neutralized  as  fast  as  formed,  the  hydrolysis 
proceeds  to  completion  and  all  the  iron  is  precipitated  as  Fe(OH)3. 
In  this  case,  the  reagent  used  to  bring  about  the  exact  neutrality 
of  the  solution  is  a  suspension  of  basic  zinc  carbonate.  (Compare 
the  similar  procedure  for  removing  traces  of  iron  in  the  preparation 
of  strontium  chloride,  No.  21.) 

Materials:     white  vitriol,  ZnS04.7H20,  57  grams  =  J-  F.W. 

anhydrous  sodium  carbonate,  Na^COs,  23  grams  = 

JF.W. 
chlorine  water,   15  cc.   (must  be  fresh  and  smell 

strongly  of  chlorine). 
Apparatus:  2  2-liter  common  bottles. 
5-inch  filter  funnel. 

Procedure:  Dissolve  the  zinc  sulphate  in  1.5  liters  of  hot  water 
in  the  2-liter  bottle.  Add  the  chlorine  water,  and  stir  well. 
Dissolve  the  sodium  carbonate  in  250  cc.  of  water;  add  12  cc.  of 
this  solution  at  once  to  the  large  bottle,  stir  and  observe  the  color 
of  the  precipitate.  If  it  is  reddish  brown  (Fe(OH)3)  add  12  cc. 
more  of  the  sodium  carbonate  solution,  which  ought  to  precipitate 
white  basic  zinc  carbonate.  The  entire  precipitate  when  mixed, 
ought  to  be  of  a  light  buff  color  showing  it  to  contain  basic  zinc 
carbonate.  Hold  the  rest  of  the  Na2CO3  solution  in  reserve.  Let 
the  solution  in  the  bottle  stand  (best  over  night)  until  the  pre- 
cipitate has  settled.  Decant  the  clear  solution  through  a  large 
filter  catching  the  filtrate  in  another  2-liter  bottle.  Finally  pour 
the  sludge  on  the  filter  and  let  it  drain.  Filter  the  rest  of  the 
Na2CO3  solution  unless  it  is  clear  and  add  it  to  the  zinc  sulphate 


206  HEAVY  METALS  OF  GROUPS  I  AND  II 

solution.  The  precipitate  should  be  pure  white.  Note  the 
effervescence.  Let  the  precipitate  settle,  —  to  J  or  J  the  volume 
of  the  solution.  This  will  take  about  30  minutes.  Draw  off  the 
clear  solution  and  wash  the  remaining  precipitate  by  decantation 
until  it  is  calculated  that  it  i&  contaminated  with  less  than  0.1 
percent  of  the  soluble  sodium  sulphate  present  at  first  (see  Note  5 
(&)).  Finally,  transfer  the  slime  to  a  large,  ordinary  filter  (Note  4 
(c)),  and  allow  it  to  drain  over  night.  It  should  now  be  in  the 
form  of  a  jelly  like  cake  which  holds  its  shape.  Without  removing 
it  from  the  filter  lift  the  latter  from  the  funnel  unfold  it  without 
tearing,  spread  it  flat  on  paper  towels,  and  leave  it  on  the  steam 
table  until  the  material  is  dry.  Pulverize  the  basic  zinc  carbonate, 
and  heat  it  moderately  in  a  porcelain  dish  over  a  free  flame  until 
all  the  carbon  dioxide  has  been  driven  off.  Test  the  product 
by  wetting  0.1  gram  with  2  cc.  of  water  and  adding  a  few  drops 
of  6-n  HC1.  The  zinc  oxide  should  dissolve,  and  there  should 
be  no  trace  of  effervescence.  Put  up  the  preparation  in  a  6-inch 
cork  stoppered  test  tube. 

QUESTIONS 

1.  Why   could   not   the   precipitate   of  basic   zinc   carbonate 
have  been  advantageously  freed  from  the  solution  by  means  of  a 
suction  filter? 

2.  Which  is  more  readily  decomposed  by  heat,  calcium  car- 
bonate, or  zinc  carbonate?    Which  then  is  the  more  strongly  basic, 
calcium  oxide,  or  zinc  oxide? 

To  some  solution  of  zinc  sulphate  add  a  solution  of  sodium 
hydroxide,  drop  by  drop,  until  the  precipitate  first  formed  re- 
dissolves.  How  is  zinc  hydroxide  similar  to  aluminium  hydroxide 
in  respect  to  its  behavior  towards  strong  acids  and  strong  bases? 
Write  ionic  equations. 

PREPARATION  33 

MERCUROUS  NITRATE,  HgNOs.H^O 

Mercury,  like  copper,  will  not  dissolve  in  non-oxidizing  acids, 
but  it  does  dissolve  in  nitric  acid.  Two  oxides  of  mercury  are 
known,  Hg2O,  and  HgO,  from  which  are  derived  respectively  the 
mercurous  and  mercuric  salts.  In  order  to  obtain  the  nitrate 
corresponding  to  the  lower  oxide,  it  is  necessary  merely  to  keep 
mercury  present  in  excess  until  after  the  acid  is  exhausted. 


MERCURIC  NITRATE  207 

Materials:     mercury,  Hg,  25  grams  =  £  F.W. 

6-n  HN03,  20  cc. 
Apparatus:  50  cc.  flask. 

3-inch  evaporating  dish. 

Procedure:  Treat  25  grams  of  mercury  in  a  flask  at  the  hood  with 
20  cc.  of  6-n  HN03,  warming  gently,  until  no  further  action  takes 
place.  Allow  to  cool  until  the  flask  can  be  held  in  the  hand,  then 
pour  the  solution  away  from  any  remaining  globule  of  mercury 
into  a  small  dish,  and  leave  to  crystallize  until  the  next  day. 
Spread  the  crystals  out  on  a  filter  paper  placed  on  a  paper  towel 
and  let  them  dry  at  room  temperature.  Put  the  product  in  a 
cork  stoppered  test  tube  as  soon  as  it  is  dry. 

QUESTIONS 

1.  Treat  0.5  gram  of  the  preparation  with  10  cc.  of  cold  water. 
It  does  not  dissolve  to  give  a  clear  solution.     Note  the  character  of 
the  residue  of  basic  salt.     Add  dilute  nitric  acid  drop  by  drop, 
until  a  clear  solution  is  obtained.     Explain  why  the  presence  of  a 
little  nitric  acid  should  enable  us  to  get  a  clear  solution. 

2.  The  cold  dilute  nitric  acid  in  Question  1  does  not  oxidize 
the  mercurous  salt.     Now  add  cold  dilute  hydrochloric  acid,  drop 
by  drop,  until  all  of  the  mercurous  salt  is  precipitated  as  white 
mercurous  chloride,  HgCl.     Look  up  the  solubility  of  mercurous 
and  mercuric  chlorides. 

3.  Filter   off   the   precipitate   obtained   in   Question   2.    The 
solution  will  contain  any  mercuric  salt  which  was  present  in  the 
sample  originally  taken.     What  reagent  could  you  use  to  test  for 
mercuric  salt  in  this  solution?     Make  the  test  and  report  the  result. 

4.  Addition  of  hydrochloric  acid  to  the  solution  containing 
nitric  acid,  gives  the  strongly  oxidizing  mixture  known  as  aqua 
regia.     What  is  the  chief  reason  why,  if  the  solution  is  cold  and 
dilute,  the  mercurous  salt  escapes  oxidation? 

PREPARATION  34 
MERCURIC  NITRATE 

When  mercury  is  heated  with  an  excess  of  nitric  acid,  mercuric 
nitrate  is  produced.  This  salt  is  exceedingly  soluble  in  water, 
and  it  can  only  be  crystallized  with  a  good  deal  of  difficulty.  When 


208  HEAVY  METALS  OF  GROUPS  I  AND  II 

a  solution  of  it  containing  an  excess  of  nitric  acid  is  evaporated, 
it  becomes  a  thick,  heavy  sirup,  which  by  further  driving  off  of 
nitric  acid  and  water  becomes  a  pasty  mass,  due  to  formation  of 
small  crystals  of  basic  nitrate,  Hg  <  Qjj3-  If  the  materials  taken 

for  the  preparation  of  this  saft  are  pure,  the  product  can  contain 
no  other  foreign  matter  than  an  excess  of  nitric  acid ;  consequently, 
in  view  of  the  difficulty  of  obtaining  good  crystals,  it  is  convenient 
to  preserve  the  salt  in  this  pasty  condition. 
Materials:     mercury,  Hg,  25  grams  =  |  F.W. 

6-normal  HN03,  60  cc. 
Apparatus:  50  cc.  flask. 

100  cc.  casserole. 

25  cc.  glass  stoppered  sample  bottle. 

Procedure:  Heat  25  grams  of  mercury  in  a  flask  with  60  cc. 
6-n  HNOs  until  it  is  all  dissolved.  Test  a  drop  of  the  solution 
by  adding  to  it  in  a  test  tube  |  cc.  of  cold  water  and  a  drop  of  dilute 
hydrochloric  acid.  A  precipitate  will  probably  form,  in  which 
case,  add  10  cc.  of  concentrated  nitric  acid  to  the  flask  and  boil 
until  a  precipitate  is  no  longer  obtained  when  tested  as  above. 
Pour  the  solution  into  a  casserole  and  evaporate  over  a  very  small 
free  flame  until  the  liquid  has  assumed  a  sirupy  consistence  and 
crystals  just  commence  to  form  on  the  surface.  Then  remove  the 
whole  mass  to  the  sample  bottle,  which  has  previously  been 
weighed;  let  it  cool  and  stopper  the  bottle. 

QUESTIONS 

1.  To  prepare  a  solution  of  this  salt  for  use  as  a  laboratory 
reagent,  explain  why  it  is  necessary  to  add  nitric  acid.     (Compare 
with  Question  1  under  Mercurous  Nitrate.) 

2.  To  a  solution  of  mercuric  nitrate  add  a  little  hydrochloric 
acid.     Now  add  a  little  stannous  chloride  solution.     What  is  the 
precipitate,  and  what  change  in  the  valence  of  mercury  must  have 
occurred  before  it  could  form? 

PREPARATION  35 
MERCURIC  SULPHOCYANATE 

In  most  of  its  properties  the  sulphocyanate  radical  resembles 
the  halogens,  with  which  it  is  often  classed,  in  the  same  manner 
that  the  ammonium  radical,  NH4,  is  classed  with  the  alkali  metals. 


MERCURIC  SULPHOCYANATE  209 

Thus  potassium  sulphocyanate,  KSCN,  yields  the  ion  SCN~  just 
as  potassium  chloride  yields  the  ion  Cl~.  Mercuric  sulphocyanate 
is  insoluble  in  water,  and  may  be  produced  by  bringing  together 
equivalent  quantities  of  solutions  of  mercuric  nitrate  and  potassium 
sulphocyanate,  but  if  an  excess  of  either  of  these  reagents  is  used, 
the  precipitate  dissolves  in  it.  From  a  consideration  of  the 
principle  of  solubility  product  one  would  predict  that  an  excess 
of  either  ion  would  cause  a  decrease  in  the  solubility  of  the  salt 
Hg(SCN)2,  but  this  effect  is  outweighed  by  the  tendency  of 
Hg(SCN)2  to  combine  with  either  Hg++  or  SCN~-ions  to  form 
fairly  stable  complex  ions  (see  page  99).  A  very  neat  expedient 
may  be  adopted  in  this  preparation  to  show  when  the  proper 
amount  of  reagent  has  been  added,  as  follows:  ferric  sulpho- 
cyanate, Fe(SCN)3,  is  a  soluble  substance  which  has  an  intensely 
red  color.  If  to  a  given  solution  of  mercuric  nitrate  a  few  drops 
of  a  ferric  salt  solution  are  added,  and  then  to  this  is  gradually 
added  a  solution  of  potassium  sulphocyanate,  the  SCN~-ions 
will  unite  with  the  Hg++-ions  so  long  as  any  of  the  latter  are 
present,  the  solution  remaining  colorless  and  the  precipitate 
Hg(SCN)2  forming  towards  the  end;  but  so  soon  as  the  Hg++-ions 
are  exhausted,  then  the  SCN~-ions  unite  with  Fe+++-ions,  pro- 
ducing the  red  compound,  and  the  appearance  of  the  red  color 
is  the  indication  to  stop. 

Mercuric  sulphocyanate  has  the  peculiar  property  that  when 
ignited  it  burns  with  the  production  of  a  very  voluminous  coherent 
ash,  which,  from  the  form  which  it  assumes,  is  called  "  Pharaoh's 
Serpent."  It  should  not  be  burned  in  an  open  room  because  of 
the  production  of  poisonous  vapors. 

Materials:     mercuric  nitrate  from  preceding  preparation. 

potassium  sulphocyanate,  KSCN,  25  grams. 

ferric  chloride  for  indicator. 
Apparatus:  2-liter  common  bottle. 

suction  filter. 

Procedure:  Dissolve  the  mercuric  nitrate  in  1  liter  of  water, 
adding  enough  nitric  acid  to  prevent  the  formation  of  any  basic 
salt.  To  this  add  10  drops  of  a  ferric  chloride  solution;  then  add 
gradually,  with  constant  stirring,  a  solution  of  the  potassium 
sulphocyanate  in  500  cc.  of  water  until  a  red  color  appears  and 


210  HEAVY  METALS  OF  GROUPS  I  AND  II 

persists  after  stirring.     Collect  the  precipitate  on  a  suction  filter, 
and  dry  it  on  paper  towels. 

The  dried  salt  may  be  made  into  the  so-called  Pharaoh's  serpent 
eggs  by  mixing  it  with  1.5  grams  of  dextrine  and  water  to  obtain 
a  paste,  placing  the*  latter  in^conical  moulds  about  1  cm.  wide 
and  1  cm.  deep,  and  letting  it  dry  out  and  harden, 

QUESTIONS 

1.  Find  out  what  is  the  degree  of  ionization  of  the  soluble 
halides  of  mercury,  i.e.,  HgCl2,  Hg(CN)2.     Do  these  salts  form 
in  this  respect,  any  exception  to  the  general  rule  regarding  the 
ionization  of  salts? 

2.  Describe  at  least  three  instances  which  have  previously  fallen 
under  your  observation  in  which  a  reagent  in  limited  amount  will 
give  a  precipitate,  but,  added  in  excess,  will  cause  the  precipitate 
to  redissolve. 

Experiments 

Review  in  Chap.  Ill  Exps.  15,  16,  19,  20,  and  21  and  the  dis- 
cussion of  the  basic  properties  of  metal  hydroxides,  p.  94,  and  of 
complex  ions,  p.  97.  In  Chap.  IV,  review  Exps.  16  and  22. 

1.  Stability  of  the  Carbonates,  (a)  Make  unacidified  solu- 
tions of  CuS04,  ZnSO4,  CdSC>4,  by  dissolving  1  gram  of  each 
in  about  15  cc.  of  water,  and  of  AgN03  by  dissolving  0.5  gram 
in  10  cc.  of  water.  To  each  solution  add  1-n  Na2CO3  until 
no  further  precipitate  is  formed  and  notice  very  carefully 
whether  any  bubbles  of  gas  escape.  In  those  cases  in  which 
a  gas  did  escape,  collect  the  precipitate  on  a  filter  wash  it 
thoroughly  with  water  until  the  excess  of  Na2C03  is  removed. 
This  is  accomplished  when  the  washings  from  the  filter  no 
longer  effervesce  when  HC1  is  added.  Now  pour  a  few  drops 
of  HC1  on  these  precipitates  and  note  that  they  effervesce 
when  they  dissolve. 

The  escape  of  carbon  dioxide  when  the  Na2CO3  is  added  shows 
that  the  copper  and  zinc  carbonates  hydrolyze,  but  the  further 
escape  of  carbon  dioxide  when  the  precipitates  are  treated  with 
acid  shows  that  the  hydrolysis  has  not  been  complete.  The  pre- 
cipitates then  must  consist  of  basic  carbonates,  such  as  Cu(OH)2.- 
CuC03.  This  shows  that  the  basic  character  of  the  hydroxides  is 


EXPERIMENTS  211 

weak,  but  not  so  weak  as  that  of  A1(OH)6,  because  not  even  a  basic 
carbonate  of  aluminum  can  be  formed.  Since  no  effervescence 
took  place  with  the  silver  salt,  the  precipitate  must  have  been  the 
neutral  carbonate,  Ag2C03,  which  indicates  that  silver  oxide  is 
more  strongly  basic  than  the  others. 

(6)  Heat  a  little  dry  basic  copper  carbonate  by  shaking  it 
in  a  test  tube  at  some  distance  above  a  small  flame.  The 
light  blue  powder  is  quickly  changed  to  black  and  the  seeth- 
ing of  the  dry  powder  shows  that  a  gas  is  being  expelled. 
A  drop  of  lime  water  is  clouded  by  the  gas.  To  the  residue 
after  it  has  cooled  add  5  cc.  of  water  and  then  a  little  HC1  and 
note  that  the  black  powder  dissolves  without  effervescence. 

The  ease  with  which  copper  carbonate  is  decomposed  by  heat 
shows  further  the  weakness  of  the  basic  character  of  copper  oxide. 

2.  Hydrolysis  of  Salts.  Recall  the  fact,  or  observe  by  experi- 
ment, that  the  salts  HgNO3,  Hg(NO3)2,  and  ZnCl2,  do  not  dissolve 
in  pure  water  to  give  a  clear  solution  but  that  a  flocculent  residue 
of  basic  salt,  HgN03.HgOH,  HgOHNO3,  ZnOHCl,  is  left.  The 
formation  of  the  basic  salt  is  a  partial  hydrolysis,  it  leaves  the 
solution  faintly  acidic;  but,  that  the  hydrolysis  is  not  extensive,  is 
shown  by  the  fact  that  a  moderate  amount  of  the  corresponding 
acid  will  in  each  case  prevent  the  formation  and  precipitation  of 
the  basic  salt. 

3.  Hydroxides.  In  separate  test  tubes  place  (a)  1  cc.  of 
1-n  CuSO4;  (6)  10  cc.  of  .05-n  AgN03;  (c)  1  cc.  of  1-n  ZnS04; 
(d)  1  cc.  of  1-n  CdCl2;  (e)  5  cc.  of  0.1-n  HgNO3;  and  (/)  5  cc. 
of  0.2-n  Hg(N03)2.  To  each  tube  add  water  to  make  a  volume 
of  10  cc.  and  then  add  6-n  NaOH  one  drop  at  a  time,  shaking 
after  each  drop.  Finally  add  in  all  5  cc.  of  the  NaOH. 

In  every  case  a  precipitate  is  formed  with  a  small  amount  of  the 
reagent  as  follows:  (a)  light  blue  Cu(OH)2,  (b)  brown  Ag2O; 
(c)  white  Zn(OH)2,  (d)  white  Cd(OH)2,  (e)  black  Hg2O,  (/)  yeUow 
HgO.  Those  precipitates  whose  formulas  are  given  as  hydroxides 
are  in  fact  likely  to  come  down  as  rather  indefinite  basic  salts, 
that  is,  as  mixtures  of  the  hydroxides  and  the  salt,  but  for  the  sake 
of  simplicity  in  discussion  it  is  allowable  to  regard  them  as  hydrox- 
ides. Silver,  mercurous  and  mercuric  hydroxides  do  not  exist; 
they  lose  water  and  form  the  oxides. 


212  HEAVY   METALS  OF  GROUPS  I  AND  II 

Of  the  above  precipitates,  zinc  hydroxide  alone  dissolves  freely 
in  excess  of  the  reagent.  It  forms  the  soluble  salt  sodium  zincate, 
Na2ZnO2,  and  it  is  thus  an  amphoteric  substance  like  aluminum 
hydroxide  (compare  Exps.  2,  3,  4,  and  5  in  Chap.  VI). 

The  copper  hydroxide  does  -not  dissolve  altogether  in  the  excess 
of  the  reagent,  but  the  solution  acquires  a  deep  blue  color  which  is 
seen  better  after  the  precipitate  settles  out.  This  color  shows  that 
there  must  be  copper  in  solution,  and  that  -the  copper  hydroxide 
possesses  amphoteric  properties  to  a  slight  degree. 

4.  Basic  Strength  of  Silver  Oxide.     Collect  on  a  filter  the 
silver  oxide  precipitate  obtained  in  (6)  of  the  last  experiment. 
Wash  it  thoroughly  with  hot  water.     Put  part  of  the  moist 
residue  on  a  piece  of  red  litmus  paper  and  note  that  the  latter 
is  turned  blue.     Pour  2  cc.  of  water  over  the  rest  of  the  brown 
residue  let  it  run  through  the  filter  and  test  the  filtrate  for 
Ag+-ions  by  adding  a  drop  of  HC1.     A  distinct  test  is  obtained. 

Test  a  silver  nitrate  or  a  silver  sulphate  solution  with  litmus 
and  note  that  the  indicator  is  not  affected . 

The  brown  silver  oxide  must  combine  with  water  to  form  the 
hydroxide  when  it  dissolves  because  of  the  tests  for  Ag+-  and  OH~- 
ions  that  are  obtained  in  the  solution.  Silver  oxide  is  thus  shown 
to  be  appreciably  soluble  and  markedly  basic.  As  would  be 
expected  of  salts  of  such  a  base,  we  find  that  silver  nitrate  and 
silver  sulphate  are  not  hydrolyzed.  Silver  oxide  is  exceptional  for 
a  heavy  metal  oxide  in  displaying  so  marked  a  basic  strength. 

5.  Ammoniates.     Repeat  Exp.  3  using  6-n  NH4OH  instead 
of  NaOH.     Note  that  in  every  case  a  limited  amount  of  re- 
agent produces  the  same  precipitate  as  NaOH,  and  that  excess 
of  the  reagent  redissolves  all  of  the  precipitates  except  those 
of  the  mercury  oxides. 

Test  the  resulting  ammonia  silver  salt  solution  for  Ag+-ions 
by  adding  a  drop  of  KC1  and  note  that  no  precipitate  is  formed. 

The  hydroxides  dissolve  in  NH4OH,  not  because  of  any  acidic 
character  but  because  of  the  ability  of  the  metal  ions  to  form 
ammoniates  (see  pp.  97-99).  Addition  of  ammonia  to  the  simple 
metal  radical  seems  to  strengthen  its  metallic  character  so  that  it 
can  exist  more  easily  as  a  positive  ion.  Thus  the  hydroxides  of 


EXPERIMENTS  213 

the  ammonio-metal  radicals,  excepting  those  of  mercury,  as  are 
soluble  and  as  highly  ionized  as  the  hydroxides  of  the  alkali  metals. 

6.  Complex  Negative  Ions,  (a)  To  1  cc.  1-n  CuSO4  add 
10  cc.  water  and  2  cc.  of  1-n  KI.  Note  that  the  solution  turns 
brown  and  that  a  precipitate  is  formed.  Let  the  precipitate 
settle,  pour  off  the  brown  solution  and  note  that  a  drop  of  it 
will  turn  some  starch  paste  blue.  Wash  the  precipitate  by 
decantation  and  note  that  it  is  white  after  the  brown  solution 
is  removed.  To  the  precipitate  suspended  in  about  2  cc.  of 
water  add  small  crystals  of  potassium  iodide.  This  dissolves 
quickly  giving  a  fairly  concentrated  solution.  Note  that  the 
precipitate  dissolves  in  the  KI  solution. 

The  ions  of  cupric  iodide  are  brought  together  but  cupric  iodide 
is  unstable  and  decomposes  into  insoluble  cuprous  iodide,  Cul,  and 
free  iodine.  With  excess  of  KI  the  soluble  salt  KCul2,  which  is 
ionized  as  K+  CuI2~,  is  formed.  Cupric  chloride,  CuCl2,  and 
cupric  bromide,  CuBr2,  do  not  decompose  in  the  same  way  into 
cuprous  salts.  (Compare  the  relative  reducing  action  of  chloride, 
bromide,  and  iodide,  Chap.  IV,  Exps.  12,  13,  and  14.) 

(6)  To  1  cc.  of  0.2-n  Hg(NO3)2  add  a  few  drops  of  1-n  KI 
and  note  a  bright  red  precipitate.  Add  a  little  more  KI  and 
note  that  the  precipitate  redissolves. 

(c)  To  1  cc.  of  0.05-n  AgN03  add  a  few  drops  of  a  saturated 
NaCl  solution.  Note  the  white  precipitate.  Add  10  cc.  of 
saturated  NaCl,  shake,  and  note  that  the  precipitate  cannot 
be  seen  to  dissolve.  To  show  that  some  does  dissolve,  filter, 
and  add  a  large  amount  of  water  to  the  clear  filtrate,  noting 
an  opalescent  precipitate. 

Mercuric  iodide,  HgI2,  is  very  insoluble  but  with  excess  iodide 
ions  it  readily  forms  the  complex  ion  HgI4  .  Similarly  silver 
chloride  can  in  presence  of  a  high  concentration  of  chloride  ions 
form  to  a  limited  extent  the  complex  ion  AgCl2~.  The  latter  is 
very  unstable,  however,  and  dilution  reduces  the  Cl~-ion  concen- 
tration sufficiently  to  allow  the  complex  ion  to  dissociate,  AgCl2~ 
^±  AgCl  I  +  Cl~,  with  a  reprecipitation  of  silver  chloride. 

The  above  experiments  show  the  strong  tendency  of  the  heavy 
metals  of  this  chapter  to  enter  into  the  formation  of  complex 
negative  ions.  (See  p.  99.)  The  negative  radical  ions  show  differ- 


214  HEAVY   METALS  OF  GROUPS   I  AND  II 

ing  tendencies  to  enter  into  these  complexes  and  in  the  decreasing 
order  of  the  strength  of  this  tendency  are  CN~,  SCN~,  I~,  Cl~. 
Sulphate  and  nitrate  ions  show  very  little  tendency  to  join  heavy 
metal  ions  in  the  formation  of  complexes  of  this  kind,  although 
it  may  be  that  the  formation' of  the  crystallized  double  salts  like 
the  alums  and  the  double  sulphate  of  potassium  and  copper  are 
due  to  some  such  cause. 

7.  Sulphides,  (a)  To  separate  tubes  containing  diluted 
heavy  metal  salt  solutions  as  in  Exp.  3,  (a)  to  (/),  add  2  cc. 
of  6-n  (NH4)2S  in  each  case  and  note  the  character  of  the  pre- 
cipitate. Collect  each  precipitate  on  a  filter,  wash  it  with  hot 
water,  and  then  pour  over  it  a  few  cc.  of  6-n  HC1. 

The  precipitates  are  CuS,  dark  brown;  Ag2S,  black;  ZnS,  white; 
CdS,  yellow;  HgS  +  Hg,  black;  and  HgS,  black.  Of  these  only 
zinc  sulphide  and  cadmium  sulphide  dissolve  in  6-n  HC1. 

(b)  To  1  cc.  of  1-n  ZnS04  add  10  cc.  of  water  and  1  cc.  of 
6-n  HC1  and  pass  hydrogen  sulphide  in  until  the  solution  is 
saturated  with  the  gas.  No  precipitate  is  formed.  Add  10 
cc.  of  1-n  NaAc  and  observe  the  white  precipitate. 

Zinc  sulphide  does  not  precipitate  from  an  acidified  solution 
because  the  S  -ion  concentration  is  repressed  by  the  H+-ions  of 
the  strong  acid,  H2S  ^±  2H+  +  S — ,  and  the  solubility  product  of 
zinc  sulphide  cannot  be  reached.  Acetate  ions  however  remove 
H+  ions,  and,  the  hydrogen  sulphide  thus  being  allowed  to  ionize 
to  a  greater  extent,  the  solubility  product  of  zinc  sulphide  is  ex- 
ceeded and  the  white  precipitate  appears  (see  Solubility  product, 
p.  Ill,  and  Exp.  22,  p.  154). 

8.  Electromotive  Series.  Review  Exp.  20,  Chap.  Ill,  and  make 
what  further  experiments  of  a  similar  nature  are  necessary  to  de- 
termine the  relative  position  in  the  electromotive  series  of  the 
heavy  metals  considered  in  this  chapter. 

GENERAL  QUESTIONS   VII 
HEAVY   METALS   OF   GROUPS   I  AND   II 

1.  Make  a  table  as  follows.  In  column  1  give  the  symbols  of 
the  metals,  copper  (leave  two  lines),  silver,  gold  (two  lines),  zinc, 
cadmium  and  mercury  (two  lines);  in  column  2,  the  formulas  of 


GENERAL  QUESTIONS  VII  215 

the  chlorides;  in  column  3,  the  solubility  of  the  chloride,  speci- 
fying sol  =  soluble,  ins  =  insoluble;  in  column  4,  the  formula  of 
the  oxide  corresponding  to  the  chloride  in  column  2 ;  in  column  5, 
the  formula  of  the  corresponding  nitrate,  if  one  exists;  in  column  6, 
the  degree  of  hydrolysis  of  the  chloride  or  nitrate,  whichever  is 
soluble,  specifying,  none,  little,  large. 

2.  Give  all  available  information  as  to  the  stability  of  the  car- 
bonates of  the  metals  of  this  chapter.     Compare  the  base  forming 
properties  of  these  metals  among  themselves,  and  also  with  the 
alkali  metals,  the  alkaline  earth  metals,  and  aluminum. 

3.  What  is  an  ammoniate?    Give  the  formulas  of  the  ammonio- 
ions  of  copper,  silver,  zinc,  and  cadmium.     How  can  crystallized 
ammonio-copper  sulphate  be  prepared?    How  would  you  make  a 
solution  of  ammonio-copper  hydroxide?    What   is  the   alkaline 
strength  of  such  a  solution? 

4.  Discuss  the  tendency  of  the  heavy  metals  of  Groups  I  and  II 
to  enter  complex  negative  ions;  give  several  examples,  at  least  one 
for  each  metal,  also  examples  in  which  the  simple  cyanide,  thio- 
cyanate,  iodide,  and  chloride  ions  are  involved.     Describe  experi- 
mental facts  to  illustrate  the  great  stability  of  potassium  argenti- 
cyanide,  KAg(CN)2,  and  the  instability  of  sodium  argentichloride, 
NaAgCl2. 

Compare  the  metals  of  this  chapter  among  themselves  with 
regard  to  their  position  in  the  electromotive  series,  also  with  the 
metals  of  the  A  families  of  Groups  I  and  II,  and  with  aluminum. 


-       CHAPTER  VIII 
THE  OXY-ACIDS  AND   SALTS  OF  THE  NON-METALS 

The  elements  which  are  distinctly  and  invariably  non-metallic 
in  character  are  boron  in  the  third  group,  carbon  and  silicon  in  the 
fourth  group,  nitrogen  and  phosphorus  in  the  fifth  group,  oxygen, 
sulphur  and  selenium  in  the  sixth  group,  and  fluorine,  chlorine, 
bromine  and  iodine  in  the  seventh  group.  Non-metallic  elements 
have  already  been  studied  in  Chapter  IV,  insofar  as  they  enter 
binary  compounds,  in  which  they  act  as  the  negative  constituents. 

When  two  non-metallic  elements  unite,  the  one  which  is  the 
least  strongly  non-metallic  is  regarded  as  the  positive  constituent ; 
it  is,  so  to  speak,  compelled  to  play  the  positive  role  in  the  com- 
pound. For  example,  sulphur  is  regarded  as  the  positive  constit- 
uent of  sulphur  trioxide.  Although  in  such  a  compound  the 
primary  valence  of  each  element  seems  to  be  satisfied,  —  sulphur, 
+  VI,  and  oxygen,  —  II,  in  sulphur  trioxide,  —  there  must  be  a 
large  residual  combining  power  (secondary  valence)  because  the 
compound  combines  so  readily  with  other  saturated  compounds. 
The  oxides  of  non-metals  thus  unite  with  metal  oxides  to  form  salts, 
and  with  water  to  form  acids,  for  example,  SO3  +  Na^O  — •»  Na2S04; 
S03  +  H2O  — >  H2SC>4.  When  such  salts  or  acids  ionize  the  non- 
metal  always  appears  as  a  constituent  of  the  complex  negative  ion. 
In  the  sulphate  ion  the  combination  of  one  sulphur,  with  a  valence 
of  +  VI,  and  four  oxygens,  with  a  valence  of  —  II  each,  would 
leave  the  unbalanced  valence  of  —  II  for  the  whole  ion;  this  is  the 
actual  valence  of  the  ion.  Hence  we  conclude  that  elements 
present  in  molecular  compounds  and  in  complex  ions  still  possess 
the  same  primary  valence  as  in  the  simple  compounds. 

Although  all  the  non-metals,  except  oxygen  and  fluorine  show 
positive  primary  valences  in  some  of  their  compounds,  nevertheless 
it  is  the  most  characteristic  property  of  the  non-metals  that  they 
do  not  form  positive  ions.  In  ions  they  are  always  combined  with 
enough  of  an  electro-negative  element  to  furnish  a  surplus  of  nega- 
tive charges  for  the  whole  ion. 

An  inspection  of  the  periodic  arrangement  of  the  elements  shows 

216 


POTASSIUM  BROMATE  AND  POTASSIUM   BROMIDE      217 

that  non-metals  do  not  occur  at  all  in  the  first  and  second  groups; 
that  they  occur  only  at  the  top  in  the  third,  fourth,  and  fifth 
groups;  and  that  in  the  sixth  and  seventh  groups  they  comprise 
all  the  members  of  the  B  families.  It  is  true  in  these  families,  as 
might  be  expected  by  recalling  characteristics  of  preceding  groups, 
that  the  strength  of  the  non-metallic  character  grows  weaker  and 
that  the  approach  towards  metallic  character  grows  more  evident, 
as  the  atomic  weight  increases;  indeed  it  is  probable  that  if  the 
elements  which  should  fit  into  the  places  below  tellurium  and 
iodine,  respectively,  are  ever  found,  they  will  display  rather 
marked  metallic  properties. 

The  characteristic  valences  of  the  sixth  and  seventh  groups  are 
VI  and  VII,  respectively,  and  the  corresponding  oxides  are  E03 
and  E207.  In  these  oxides  and  in  the  compounds  derived  from 
them,  there  is  little  dissimilarity  between  the  A  and  B  families. 
Thus  perchlorates  and  permanganates  are  in  every  way  analogous 
to  each  other,  as  are  also  sulphates  and  chromates.  In  the  lower 
states  of  valence,  the  elements  of  the  B  families  are  entirely  dif- 
ferent from  those  of  the  A  families,  the  B  family  elements  forming 
exclusively  negative  ions,  S — ,  SO2 — ,  whereas  the  A  family 
elements  form  positive  ions,  Cr++,  Cr+++. 


PEEPARATION  36 
POTASSIUM  BROMATE  AND  POTASSIUM  BROMIDE 

Bromine,  like  chlorine,  hydrolyzes  to  a  considerable  extent  ac- 
cording to  the  reversible  reaction : 

Br2  +  H20  ^±  HBr  +  HBrO.  (1) 

In  presence  of  a  base  both  acids  are  neutralized  as  fast  as  formed 
and  reaction  (1)  proceeds  to  completion.  Thus  the  complete 
reaction  in  presence  of  KOH  is: 

Br2  +  2KOH  -»  KBr  +  KBrO  +  H2O.  (2) 

Hypobromites  are  unstable,  and,  at  boiling  temperature,  and 
particularly  in  slightly  acid  solution,  undergo  an  action  of  oxidation 
and  reduction,  the  total  result  of  which  is : 

3KBrO  ->  KBrO3  +  2KBr.  (3) 


218         OXY-ACIDS  AND  SALTS  OF  THE   NON-METALS 

Multiplying  equation  (2)  by  3  and  adding  equation  (3)  the  equa- 
tion 

3Br2  +  6KOH  ->  KBrO3  +  5KBr  +  3H2O  (4) 

is  obtained.  This  equation  represents  the  sum  total  of  the  chemi- 
cal changes  occurring  in  the  process,  and  on  it  the  calculation  of 
quantities  of  materials  and  products  is  to  be  based. 

The  reaction  represented  in  (3)  does  not  take  place  to  an  appre- 
ciable extent  in  alkaline  or  neutral  solution,  but  when  the  neutral 
point  is  overateppped  by  a  mere  trace  of  acid  the  reaction  readily 
completes  itself  in  a  few  minutes  at  the  boiling  temperature.  A 
little  excess  of  bromine,  shown  by  a  reddish  color,  gives,  in  virtue 
of  the  hydrolysis  above  mentioned,  the  requisite  degree  of  acidity. 
The  following  train  of  equations  gives  an  idea  of  the  way  in  which 
this  trace  of  bromine  promotes  the  reaction.  It  will  be  noticed 
that  in  equation  (9)  the  bromine  used  in  equation  (5)  is  regener- 
ated, so  that  a  trace  of  bromine  working  over  and  over  again 
suffices  to  convert  all  of  the  hypobromite  into  bromate.  The  addi- 
tion of  all  of  the  equations  (5)  to  (9)  in  the  train  gives  the  equation 
(10)  for  the  total  net  change  which  is  identical  to  equation  (3) 

Br2  +  H2O  ->  HBr  +  HBrO  (5) 

HBr  +  KBrO  -*  KBr  +  HBrO  (6) 

2HBrO  +  KBrO  -»  KBrO3  +  2HBr  (7) 

HBr  +  KBrO  ->  KBr  +  HBrO  (8) 

HBrO  +  HBr  -»  Br2  +  H20  (9) 

3KBrO  ->  KBrO3  +  2KBr  (10) 

Materials:  potassium  hydroxide:  1  F.W.  =  28  grams;  this 
material  should  be  as  nearly  free  from  carbonate 
as  possible.  Since  it  is  very  deliquescent  it  will 
doubtless  contain  water  and  a  somewhat  greater 
weight  than  28  grams  will  have  to  be  taken. 
Take  31  grams  if  the  analysis  of  the  material  is 
not  known. 

bromine:  Br2,  j  F.W.  =  40  grams  =  12^  cc. 
Apparatus:  300  cc.  flask. 

30  cc.  iron  crucible. 

suction  filter. 

small  ordinary  filter. 

4-inch  evaporating  dish. 

graduate. 


POTASSIUM   BROMATE  AND  POTASSIUM   BROMIDE      219 

Procedure:  Dissolve  the  potassium  hydroxide  in  100  cc.  of  water 
in  a  300  cc.  Erlenmeyer  flask.  At  the  hood  obtain  the  liquid 
bromine  in  a  measuring  cylinder  and  perform  all  the  operations 
with  bromine  under  the  hood.  Cool  the  KOH  solution  to  room 
temperature  and  pour  the  bromine  into  it,  about  1  cc.  at  a  time, 
rotating  the  contents  of  the  flask  until  the  bromine  has  dissolved 
after  each  addition.  When  all  the  bromine  is  added,  it  should  be 
in  slight  excess,  which  is  shown  by  a  distinct  reddish  tint  in  the 
solution,  not  merely  a  yellow  color. 

Now  heat  the  contents  of  the  flask  to  boiling  and  boil  until  the 
excess  of  bromine  has  been  expelled.  Then  cool  to  15°  or  lower. 
Collect  the  crystals  on  a  suction  filter.  Preserve  filtrate  to  obtain 
by-product. 

Dissolve  crystals  in  four  times  their  weight  of  hot  water.  Unless 
the  solution  is  perfectly  clear,  filter  it  hot,  and  without  suc- 
tion, to  remove  dirt,  rinsing  the  filter  with  about  5  cc.  of  boiling 
water. 

Cool  the  filtrate  to  below  15°  as  before,  and  collect  crystals. 
Add  mother  liquor  to  that  reserved  for  obtaining  the  by-product, 
potassium  bromide. 

Dissolve  .05  gram  of  crystals  in  2  cc.  hot  water,  add  1  drop  of 
AgNOs  solution.  A  precipitate  while  the  solution  is  at  boiling 
temperature  is  silver  bromide  and  shows  that  the  product  has  not 
been  purified  from  bromide.  Repeat  the  recrystallization  as  many 
times  as  necessary  to  obtain  a  pure  product. 

Potassium  Bromide.  —  Combine  all  of  the  mother  liquors, 
evaporate  in  a  porcelain  dish  until  a  pasty  mass  is  obtained,  mix 
this  thoroughly  with  5  grams  of  powdered  charcoal,  and  dry  the 
mass  completely.  Pulverize  the  dry  mixture  in  a  mortar  and  heat 
it  to  redness,  for  20  minutes,  in  an  iron  crucible  surrounded  by  an 
asbestos  mantle.  Extract  the  product  with  60  cc.  hot  water, 
filter,  wash  the  residue  and  filter  with  an  additional  15  cc.  of  hot 
water,  and  evaporate  the  solution  to  dryness  to  obtain  potassium 
bromide.  The  solution  of  potassium  bromide  "  creeps."  If  it  has 
to  be  set  away  over  night  the  vessel  containing  it  should  be  placed 
in  a  clean  large  dish  to  catch  any  of  the  salt  that  creeps  over  the 
edge  of  the  smaller  vessel. 

Test  the  product  for  absence  of  bromate  by  dissolving  some  in  a 
little  water  and  acidifying  with  sulphuric  acid.  If  no  bromate 


220         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

is  present,  no  free  bromine  will  be  produced,  i.e.,  the  solution  will 
remain  colorless  and  odorless. 

Put  up  the  two  products  in  cork-stoppered  test  tubes. 

QUESTIONS 

1.  When  the  bromate  is  tested  with  silver  nitrate  why  is  it 
necessary  to  have  the  solution  hot?     (Look  up  solubility  of  silver 
bromate.) 

2.  How  does  dilute  H2S04  react  with  a  pure  bromate?     Explain, 
with  reactions,  the  test  for  the  presence  of  bromate  in  the  bromide. 

3.  Explain  why  it  would  not  be  possible  to  free  a  preparation  of 
potassium  bromide  of  a  little  bromate  by  recrystallization. 

4.  Explain  why  potassium  bromate  can  be  readily  freed  of 
bromide  by  recrystallization. 

PREPARATION  37 
POTASSIUM  CHLORATE 

Chlorine  hydrolyzes  to  an  even  greater  extent  than  bromine. 
(Study  the  discussion  of  the  preceding  preparation.)  The  bleach- 
ing power  of  chlorine  is  due  directly  to  the  oxidizing  action  of  the 
hypochlorous  acid  produced  by  the  hydrolysis.  The  formation 
of  chlorate  in  this  preparation  is  also  a  result  of  the  oxidizing  action 
of  hypochlorous  acid. 

In  the  action  of  chlorine  with  an  alkali  hydroxide,  six  equivalents 
of  the  latter  must  react  in  order  to  produce  one  equivalent  of 
alkali  chlorate.  To  economize  in  potassium  hydroxide  which  is 
more  expensive  than  sodium  hydroxide  we  shall  use  one  equivalent 
only  of  the  former  and  five  equivalents  of  *the  latter.  The  five 
equivalents  of  sodium  chloride  which  could  be  recovered  as  a  by- 
product is  not  of  great  value,  and  we  shall  discard  it  in  this  prep- 
aration. 

Materials:     potassium  hydroxide,  16  grams  \  , ,       , .  ,    r  » 

,      . ,  the  stick  form  of 

sodium  hydroxide,  44  grams      J 

these  hydroxides,  as  free  of  carbonate  as  possible, 
should  be  used.  Allowing  roughly  10  percent  as 
water  gives  the  net  amounts:  KOH,  J  F.W.; 
NaOH,  li  F.W. 


POTASSIUM  CHLORATE  221 

chlorine,  to  be  generated  from  coarse  manganese 

dioxide  and  12-n  HC1. 
Apparatus:  500  cc.  flask  with  2-holed  rubber  stopper. 

chlorine  generator,  use    1500  cc.  round-bottomed 

flask, 
suction  filter. 

Procedure:  Calculate  the  amount  of  manganese  dioxide  and  12-n 
HC1  required  to  generate  the  chlorine  necessary  to  react  with  the 
alkalies.  Take  10  percent  in  excess  of  this  amount  and  fit  up  the 
chlorine  generator.  Dissolve  the  potassium  and  sodium  hydrox- 
ides together  in  70  cc.  of  water  in  the  500  cc.  flask,  but  do  not  filter 
the  solution,  even  if  it  is  not  entirely  clear.  Arrange  a  wide 
delivery  tube  to  bubble  chlorine  into  the  alkali  solution  in  the 
flask,  which  is  supported  on  a  lamp  stand  so  that  it  can  be  heated. 
The  exit  tube  from  the  flask  is  prolonged  to  a  bottle  containing  6-n 
NaOH  to  absorb  any  excess  chlorine.  Pass  chlorine  into  the  alkali 
solution  until  the  latter  is  saturated  with  it.  Let  the  reaction  heat 
the  solution.  Finally  make  sure  that  the  solution  is  saturated 
with  chlorine.  Remove  the  generator  fittings,  close  the  flask  with 
a  solid  rubber  stopper  and  shake  vigorously;  if  the  upper  part  of 
the  flask  still  contains  chlorine  gas  the  solution  is  saturated.  Boil 
the  contents  of  the  flask  gently,  avoiding  "  bumping,"  until  the 
excess  of  chlorine  is  expelled;  then  pour  it  all,  while  still  at  boiling 
temperature  onto  the  suction  filter.  Stop  the  suction  before  any 
air  has  been  drawn  into  the  layer  of  crystals;  add  15  cc.  of  water 
to  the  flask,  heat  it  to  boiling,  pour  it  onto  the  salt  crystals  in  the 
filter,  and,  after  it  has  soaked  in,  apply  the  suction.  In  this  way 
nearly  all  of  the  potassium  chlorate  is  washed  into  the  filtrate. 
Cool  the  filtrate  to  0°  and  collect  the  potassium  chlorate  crystals 
on  the  suction  filter.  Dissolve  the  moist  crystals  in  three  times 
their  weight  of  water.  If  the  solution  is  not  clear,  pour  it  through 
a  small  common  filter,  rinsing  the  filter  with  10  cc.  of  boiling  water 
to  carry  through  any  potassium  chlorate  which  crystallized  in  the 
filter.  Purify  by  recrystallization  until  the  product  is  free  from 
chloride.  Preserve  it  in  a  cork-stoppered  test  tube. 

A  product  of  15  grams  is  to  be  regarded  as  satisfactory.  The 
mother  liquors  should  all  have  been  saved  to  work  over  again  if  the 
recrystallizations  have  not  been  skilfully  enough  carried  out  the 
first  time. 


222         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

QUESTIONS 

1.  Tabulate  the  solubilities  at  high  and  low  temperatures  of 
the  salts  concerned,  and  arrange  a  tabular  outline  of  the  method 
to  be  followed  if  the  mother  liquors  are  worked  over. 

2.  The  modern  commercial  rnethod  of  making  potassium  chlo- 
rate is  by  the  electrolysis  of  a  potassium  chloride  solution.     What 
are  the  primary  products  formed  at  the  two  electrodes?     Explain 
how,  when  the  primary  products  are  allowed  to  mix  in  the  cell, 
the  reactions  are  similar  to  those  of  this  and  the  preceding  prep- 
arations. 

PKEPARATION  38 
POTASSIUM  IODATE 

As  is  well  known,  the  chemical  affinity  of  the  halogens  for 
hydrogen  or  positive  elements  decreases  in  passing  from  fluorine 
to  iodine;  but  the  affinity  for  oxygen  increases  in  this  order,  so  that 
iodates  and  iodic  acid  (IzO*,)  are  more  stable  than  chlorates  and 
chloric  acid  (C1205).  Use  is  made  of  this  fact  in  the  following 
preparation,  in  which  the  total  change  is  represented  fairly  closely 
by  the  equation, 

KC1O3  +  I  ->  KIO3  +  Cl. 

The  actual  reaction,  however,  is  not  so  simple  as  this.  The 
presence  of  a  small  amount  of  acid  is  necessary  to  make  it  take 
place.  This  acid  gives  rise  to  a  little  free  chloric  acid,  which  is  a 
far  stronger  oxidizing  agent  than  potassium  chlorate,  and  oxidizes 
iodine  to  iodic  acid.  The  latter  acid  reacts  with  more  potassium 
chlorate  and  thus  chloric  acid  is  regenerated.  It  will  be  noticed 
that  in  carrying  out  the  following  directions  more  iodine  is  taken 
than  is  necessary  to  react  with  the  potassium  chlorate  according 
to  the  equation  given  above.  This  excess  of  iodine  is  oxidized  to 
iodic  acid  by  a  part  of  the  free  chlorine  which  is  represented  in  the 
equation  as  escaping. 

Materials:     potassium  chlorate,  KC1O3,  31  grams  =  J  F.W. 

iodine,  36  grams. 
Apparatus:  800  cc.  flask. 

short  stemmed  funnel. 

pan  of  cold  water. 

suction  filter. 


IODIC  ACID;    IODINE  PENTOXIDE  223 

Procedure:  Dissolve  31  grams  of  potassium  chlorate  by  warming 
it  with  100  cc.  of  water  in  an  800  cc.  flask.  Add  36  grams  of  pow- 
dered iodine  and  hang  a  small  funnel  in  the  neck  of  the  flask  to 
prevent,  to  some  extent,  the  escape  of  iodine  vapor.  Place  a  pan 
of  cold  water  close  at  hand;  then  add  1  cc.  of  6-n  nitric  acid  to  the 
flask,  and  warm  rather  carefully  until  a  brisk  reaction  commences. 
Then  allow  the  reaction  to  proceed  so  that  violet  vapors  fill  the 
flask,  but  if  iodine  starts  to  escape  through  the  funnel,  check  the 
reaction  by  dipping  the  flask  for  a  moment  in  the  cold  water. 
After  the  reaction  has  moderated  warm  the  solution  until  the 
iodine  color  has  disappeared,  and  then  boil  it  for  about  10  minutes 
to  expel  most  of  the  free  chlorine.  The  solution  now  contains  a 
considerable  quantity  of  iodic  acid  in  addition  to  the  potassium 
iodate.  Add  a  solution  of  potassium  hydroxide  until  the  neutral 
point  is  just  reached  (test  by  dipping  a  stirring  rod  in  the  solution 
and  touching  it  to  litmus  paper).  Allow  the  solution  to  cool, 
collect  the  crystals  of  potassium  iodate,  and  evaporate  the  mother 
liquor  to  obtain  another  crop  of  crystals.  Purify  the  entire  product 
by  dissolving  it  in  four  times  its  weight  of  hot  water,  cooling  and 
collecting  the  crystals.  Dry  the  product  and  preserve  it  in  a  2- 
ounce  cork-stoppered  bottle. 

QUESTIONS 

1 .  Explain  the  secondary  reaction  of  the  foregoing  preparation  in 
which  chlorine  reacts  with  iodine. 

2.  From  knowledge  of  the  reactions  of  chlorine  and  bromine 
predict  how  iodine  would  react  with  a  KOH  solution.     Find  from 
reference  books  if  this  prediction  is  borne  out  by  the  facts. 

PREPARATION  39 
IODIC  ACID;    IODINE  PENTOXIDE 

Iodine  pentoxide  is  a  white  solid  substance  that  at  ordinary 
temperatures  is  entirely  stable.  It  cannot  be  prepared  by  direct 
synthesis  from  iodine  and  oxygen,  because  when  cold  the  elements 
combine  too  slowly,  and  when  heated  the  compound  is  unstable. 
It  may  be  readily  prepared  by  the  direct  oxidation  of  iodine  by 
means  of  strong  oxidizing  agents,  such  as  concentrated  nitric  acid 
or  chlorine.  One  method  for  the  oxidation  of  iodine  has  already 
been  illustrated  under  the  preparation  of  Potassium  Iodate,  but 


224         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

there  the  conditions  were  such  that  a  salt  of  iodic  acid  was  obtained 
rather  than  the  free  acid  or  its  anhydride.  Starting  with  this  salt, 
however,  the  free  acid  is  easily  obtained  by  metathetical  reactions 
which  depend  on  the  insolubility  of  barium  iodate  and  the  still 
greater  insolubility  of -barium  sulphate. 

Materials:     potassium  iodate,  KI03,  43  grams  =  £  F.  W. 

barium  nitrate,  Ba(NO3)2,  26  grams  =  -^  F.W. 

36-n  H2SO4,  8  cc. 
Apparatus:  tall  600  cc.  beaker. 

500  cc.  casserole. 

suction  filter. 

Procedure:  Dissolve  the  potassium  iodate  and  the  barium  nitrate, 
separately,  each  in  250  cc.  of  hot  water,  and  mix  the  two  solutions 
at  the  boiling  temperature  while  stirring  well.  Cool  the  mixture, 
let  the  heavy  precipitate  settle,  decant  off  the  clear  liquid,  and 
wash  the  salt  twice  by  decantation  with  pure  water.  Drain  the 
barium  iodate  on  a  suction  filter,  and  wash  it  on  the  filter  with 
cold  water.  Then  remove  it  to  a  porcelain  dish,  suspend  it  in 
250  cc.  of  water,  heat  to  boiling,  and  stir  in  a  solution  of  8  cc. 
36-n  H2SO4  in  100  cc.  of  water.  Keep  this  mixture  well  stirred  at 
the  boiling  temperature  for  at  least  10  minutes,  since  the  conver- 
sion of  solid  barium  iodate  into  solid  barium  sulphate  is  a  reaction 
that  requires  some  time.  Filter  the  solution  and  rinse  the  last 
of  the  iodic  acid  from  the  solid  barium  sulphate  by  washing  two 
or  three  times  on  the  filter  with  small  portions  of  water.  Evap- 
orate the  solution  in  a  casserole  to  a  small  volume,  and  finally, 
holding  the  casserole  in  the  hand,  keep  the  contents  rotating,  so 
that  the  whole  inside  of  the  dish  is  continually  wet,  and  evaporate 
until  solid  iodic  acid  separates  in  some  quantity.  Cool  completely 
and  rinse  the  crystals  with  three  successive  portions  of  10  cc.  each 
of  16-normal  nitric  acid,  triturating  the  crystals  thoroughly  with 
each  portion  of  the  acid.  Warm  the  casserole  carefully  until  the 
product  is  perfectly  dry  and  ceases  to  give  off  acid  vapors.  This 
warming  will  convert  the  iodic  acid  to  a  large  extent  into  the  an- 
hydride I2O5.  Place  the  iodine  pentoxide  at  once  in  a  2-ounce 
cork-stoppered  sample  bottle. 

To  obtain  completely  anhydrous  iodine  pentoxide,  the  product 
could  be  heated  for  some  time  in  an  oven  at  about  200°,  Crys- 


POTASSIUM  PERCHLORATE  225 

tallized  iodic  acid  could  be  obtained  by  dissolving  the  product 
in  a  very  little  water,  in  which  it  is  extremely  soluble,  and  allowing 
the  solution  to  evaporate  slowly. 

QUESTIONS 

1.  Dissolve  a  little  of  the  iodine  pentoxide  in  water.     Test  the 
solution  to  show  whether  it  contains  a  strong  acid.     How? 

2.  Heat  J  gram  of  iodine  pentoxide  in  a  dry  test  tube.     Insert 
a  glowing  splinter  in  the  tube.     Note  whether  the  entire  substance 
can  be  volatilized;  also  if  any  of  the  original  substance  deposits  in 
the  cooler  part  of  the  tube. 

PREPARATION  40 
POTASSIUM  PERCHLORATE 

When  potassium  chlorate  is  heated  to  about  400°  it  may  de- 
compose according  to  either  of  the  following  independent  reactions: 

4KC103  =  KC1  +  3KC104,  (1) 

KC103  =  KC1  +  IJOa.  (2) 


The  second  reaction  is  accelerated  by  catalyzers,  such  as  man- 
ganese dioxide  or  ferric  oxide,  or  in  fact  any  material  with  a  rough 
surface.  Too  high  a  temperature  also  causes  reaction  (2)  princi- 
pally to  take  place.  On  the  other  hand,  if  the  temperature  is 
maintained  at  the  right  point,  the  salt  is  free  from  dirt,  and  the 
inside  of  the  crucible  is  perfectly  clean  and  free  from  roughness, 
the  decomposition  proceeds  mainly  according  to  reaction  (1). 
Potassium  perchlorate  is  very  sparingly  soluble  in  cold  water  and 
may  be  separated  from  potassium  chloride  and  any  undecomposed 
potassium  chlorate  by  crystallization. 

Materials:     potassium  chlorate,  KC103,  61  grams  =  J  F.W. 
Apparatus:  100  cc.  porcelain  crucible. 
suction  filter. 

Procedure:  Place  61  grams  of  potassium  chlorate  in  a  dry,  clean 
100  cc.  porcelain  crucible,  the  glaze  of  which  is  in  perfect  condition. 
Place  a  small  watch  glass  over  the  crucible  to  prevent  loss  of 
particles  of  the  salt  by  decrepitation,  and  heat  gently  until  the 
charge  just  melts.  Then  remove  the  watch  glass  and  keep  the 
melt  just  hot  enough  to  maintain  a  brisk  evolution  of  oxygen, 


226         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

but  do  not  increase  the  temperature  when  the  mass  shows  a  ten- 
dency to  grow  solid.  At  the  end  of  about  20  minutes  the  melt 
should  begin  to  stiffen  around  the  edges  and  become  more  or  less 
pasty  or  semi-solid  throughout;  when  this  point  is  reached,  let 
the  contents  of  the  'crucible  cdol  completely,  then  cover  it  with 
50  cc.  of  water,  and  let  it  stand  until  it  is  entirely  disintegrated. 
Collect  the  undissolved  potassium  perchlorate  on  a  suction  filter 
and  wash  it  with  two  successive  portions  o£,15  cc.  of  cold  water 
(see  Note  5  (a)  on  page  9).  Redissolve  the  salt  in  hot  water 
(see  solubility  table)  and  allow  it  to  recrystallize.  About  30  grams 
of  potassium  perchlorate  should  be  obtained.  A  few  crystals  of 
the  product  should  give  no  yellow  color  (C1O2)  when  treated  with 
a  few  drops  of  concentrated  hydrochloric  acid.  The  product 
should  be  entirely  free  from  chloride  (test  with  silver  nitrate). 

QUESTIONS 

1.  Why  is  manganese  dioxide  added  when  oxygen  is  prepared 
by  heating  potassium  chlorate? 

2.  What  is  the  reaction  of  hydrochloric  acid  with  hypochlorous, 
chloric,  and  perchloric  acids,  respectively? 

3.  What  are  the  four  oxy-acids  of  chlorine?     Compare  their 
stability. 

4.  To  what  extent  are  hydrochloric,  hypochlorous,  chloric,  and 
perchloric  acids  dissociated  electrolytically  in  dilute  solution? 

5.  How  could  pure  perchloric  acid  be  prepared  from  potassium 
perchlorate? 

6.  What  is  the  solubility  of  silver  chlorate  and  of  silver  per- 
chlorate?   How  may  preparations  of  chlorates  and  perchlorates 
be  tested  for  the  presence  of  chlorides? 

PEEPARATION  41 
SODIUM  THIOSULPHATE,  Na2S2O3.5H2O 

Sodium  sulphite  is  a  salt  of  the  lower  oxide  of  sulphur,  and 
may  thus  be  regarded  as  unsaturated  with  respect  to  oxygen;  it 
is,  in  fact,  capable  of  slowly  absorbing  oxygen  from  the  air  and 
thereby  going  over  into  sulphate.  If  it  is  allowed  to  react  with 
sulphur,  the  latter  enters  into  the  compound  in  much  the  same 
way  as  oxygen,  and  £/wosulphate  instead  of  sulphate  is  formed. 


SODIUM  THIOSULPHATE  227 

The  sulphur  so  taken  up  certainly  plays  a  different  function  than 
the  sulphur  already  contained  in  the  compound,  although  it  is 
perhaps  a  question  whether  the  thiosulphate  is  exactly  the  same 
compound  as  sulphate,  except  that  one  oxygen  atom  is  replaced 
by  a  sulphur. 

Sodium  sulphite  is  conveniently  prepared  by  allowing  sulphur 
dioxide  (sulphurous  acid)  to  react  with  sodium  carbonate.  It  is 
practically  impossible,  however,  to  distinguish  the  exact  point  at 
which  the  normal  sulphite  (Na2SO3)  is  formed;  therefore  it  is  more 
expedient  to  divide  a  given  amount  of  sodium  carbonate  into  two 
equal  parts,  to  fully  saturate  one  part  with  sulphur  dioxide,  where- 
by sodium  bisulphite,  NaHS03,  is  formed,  and  to  add  the  other 
half  of  the  sodium  carbonate,  thereby  obtaining  the  normal 
sulphite,  Na2SO3. 

Materials:     anhydrous  sodium  carbonate,  Na^COs,  106  grams  = 
1  F.W. 

sulphur  dioxide;  this  gas  is  most  conveniently 
drawn  from  steel  cylinders  in  which  liquid  sulphur 
dioxide  is  held  under  pressure.  It  can  be  pre- 
pared by  the  action  of  copper  turnings  on  36-n 
H2S04. 

sulphur  (powdered  roll  sulphur),  48  grams. 
Apparatus:  2  350  cc.  flasks  with  2-hole  rubber  stoppers  and 
delivery  tubes. 

600  cc.  beaker  with  watch  glass  to  cover  it. 

5-inch  funnel  and  10-inch  filters. 

8-inch  porcelain  dish. 

6-inch  crystallizing  dish. 

Procedure:  Dissolve  53  grams  of  the  sodium  carbonate  in  300  cc. 
of  hot  water,  and  place  about  five-sixths  of  the  solution  in  one 
flask  and  the  remainder  in  another  flask.  Connect  these  flasks 
in  series  so  that  sulphur  dioxide  gas  may  be  passed  first  into  the 
larger  volume  of  solution,  and  what  is  there  unabsorbed  may  pas/5 
on  through  the  second  flask.  Pass  a  vigorous  stream  of  the  gas 
into  the  solutions.  After  a  short  time  a  marked  frothing  occurs 
in  the  first  flask,  due  to  the  escape  of  carbon  dioxide,  and  after 
this  frothing  ceases  a  similar  frothing  soon  commences  in  the 
second  flask.  When  the  latter  ceases,  pass  the  gas  a  little  while 
longer  until  sulphur  dioxide  escapes  freely  from  the  second  bottle. 


228         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

Then  place  the  solution  of  sodium  bisulphite  in  a  600  cc.  beaker, 
and  cautiously  add  the  remaining  53  grams  of  sodium  carbonate. 
Boil  the  solution  uncovered  for  15  minutes  replacing  water  lost 
by  evaporation;  then  add  the  sulphur,  cover  the  beaker  with  a 
watch  glass,  and  keep  the  mixture  gently  boiling  for  an  hour  and 
a  half  or  longer,  again  replacing  all  of  the  water  boiled  off.  Filter 
the  solution  without  suction.  Rinse  the  residual  sulphur  onto 
the  filter  with  15  cc.  of  water  catching  th.e  drainings  with  the 
filtrate.  Then  wash  the  sulphur,  dry  it,  and  weigh  it  to  find  if 
the  calculated  quantity  has  reacted.  Evaporate  the  filtrate  in  a 
porcelain  dish  to  200  cc.,  let  it  cool  to  30°;  if  any  unchanged 
sodium  sulphite  is  left  it  will  separate  as  a  crystal  meal  which  is  to 
be  filtered  off.  Leave  the  solution  uncovered  at  room  temperature 
in  a  6-inch  crystallizing  dish  until  an  abundant  crop  of  crystals  is 
obtained;  discard  the  final  30  cc.  of  mother  liquor.  Sodium  thio- 
sulphate  can  remain  in  highly  supersaturated  solution;  it  is  best 
therefore  to  add  some  seed  crystals  to  the  solution  set  to  crystallize. 
Wrap  the  product  in  paper  towels  and  leave  it  over  night  to  dry. 
Preserve  it  in  a  10-ounce  cork-stoppered  bottle. 


QUESTIONS 

1.  Dissolve  J  gram  of  the  product  in  5  cc.  of  water  and  add  2  cc. 
of  hydrochloric   acid.     Observe   the   odor   and   the   precipitate. 
What  is  the  free  acid  corresponding  to  the  salt,  sodium  thio- 
sulphate?    What  can  be  said  regarding  the  stability  of  this  acid? 

2.  What  is  the  valence  of  sulphur  in  each  of  the  salts,  sodium 
sulphide,  sodium  sulphite,  and  sodium  sulphate?     State  in  each 
case  whether  the  sulphur  plays  the  part  of  a  positive  or  negative 
element. 

3.  Distinguish  between  the  parts  played  by  the  two  atoms  of 
sulphur  in  sodium  thiosulphate. 

4.  When  sulphur  dioxide  was  passed  into  the  sodium  carbonate 
solution  the  following  distinct,  stages  in  the  process  were  noted: 
(a)  The  gas  passed  into  the  solution  in  distinct  bubbles  and  was 
in  large  part  absorbed.     (6)  Effervescence  took  place  with  minute 
bubbles  arising  from  every  part  of  the  solution,     (c)  Effervescence 
ceased,  and  the  gas  entered  the  solution  again  in  clear,  distinct 
bubbles,  but  still  it  was  for  the  most  part  absorbed,     (d)  The  gas 
passed  through  the  solution  in  distinct  bubbles  and  was  entirely 


EXPERIMENTS  229 

unabsorbed.  Look  up  the  degree  of  ionization  of  both  hydrogens, 
both  of  sulphurous  and  carbonic  acids.  Write  equations  for  the 
reactions  taking  place  during  each  of  the  stages  enumerated  above. 

Experiments 

1.  Hypochlorites.  (a)  Stir  5  grams  of  bleaching  powder 
with  5  cc.  of  water.  It  cannot  be  seen  to  dissolve  at  all.  Add 
1  cc.  of  the  suspension  to  100  cc.  of  water,  and  note  that  it 
imparts  a  cloudiness  to  the  whole  so  that  it  cannot  yet  be  told 
whether  any  has  dissolved.  Pour  the  rest  of  the  undiluted 
suspension  into  a  filter  and  collect  the  filtrate.  To  1  cc.  of  the 
filtrate  add  a  few  drops  of  6-n  HC1.  The  color  and  odor  of 
chlorine  are  perceived. 

Bleaching  powder  is  prepared  by  treating  slaked  lime,  Ca(OH)2, 
with  chlorine  and  it  is  essentially  a  mixture  of  equivalent  amounts 
of  calcium  chloride,  CaCl2,  and  calcium  hypochlorite,  Ca(OCl)2. 
The  formula  of  the  solid  material  is  written  CaOCl2,  indicating 
the  belief  that  it  is  not  a  mixture  of  two  salts,  but  rather  a  mixed 
salt  in  which  each  molecule  contains  a  chloride  and  a  hypochlorite 
radical. 

Calcium  chloride  and  calcium  hypochlorite  are  both  extremely 
soluble;  but  bleaching  powder  always  contains  a  considerable 
excess  of  calcium  hydroxide  which  has  not  reacted  with  the 
chlorine,  and  also  a  good  deal  of  calcium  carbonate.  It  is  for  this 
reason  that  it  does  not  appear  to  the  eye  to  dissolve.  That  hypo- 
chlorite dissolves  freely  out  of  the  bleaching  powder  is  shown  by 
the  action  with  hydrochloric  acid. 

2HC1  +  Ca(OCl)Cl  -»  CaCl2  +  H2O  +  C12. 

(6)  Soak  a  piece  of  colored  cotton  cloth  in  one  half  of  the 
filtrate  obtained  in  (a).  Different  colors  bleach  with  varying 
ease,  but  it  is  probable  that  this  one  will  decolorize  hardly 
perceptibly.  Remove  the  cloth  with  the  solution  still  cling- 
ing to  it  and  immerse  it  in  very  dilute  H2SO4  (1  cc.  of  6-n  acid 
to  50  cc.  of  water).  The  color  now  rapidly  disappears. 

Add  5  cc.  of  6-n  NaOH  to  the  other  half  of  the  filtrate  from 
the  bleaching  powder  suspension.  A  voluminous  white  precip- 
itate is  formed.  Soak  another  piece  of  cloth  in  this  sus- 
pension and  note  that  it  is  not  bleached  at  all. 


230         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

It  is  a  well  known  fact  that  dry  chlorine  does  not  bleach  dry 
cloth,  but  that  chlorine  water  bleaches  it  easily.  It  is  known  that 
chlorine  hydrolyzes  (see  Prep.  37)  and  the  bleaching  is  attributed 
to  the  hypochlorous  acid,  which,  on  account  of  its  instability,  is 
a  strong  oxidizing  agent  and *  oxidizes  the  colored  substance  to  a 

colorless  one. 

color  +  HOC1  ->  color  oxide  +  HC1. 

The  hypochlorite  solution  contains  OCl~:ions  and  the  fact  that 
this  solution  does  not  bleach  rapidly  indicates  that  the  ions  are  not 
the  principal  bleaching  agent.  Addition  of  sulphuric  acid  pro- 
duces un-ionized  hypochlorous  acid,  H+  +  OC1~  — »  HOC1,  the 
ionization  of  this  acid  in  0.1-n  solution  being  0.06  of  one  percent. 
Thus  it  is  clear  that  HOC1  must  be  the  principal  bleaching  agent. 
That  the  hypochlorite  solution  did  bleach  the  cloth  slowly  might 
have  been  due  to  a  slow  action  of  the  OCl~-ion,  or  perhaps  to  a 
small  amount  of  HOC1  produced  by  hydrolysis  of  the  salt,  or  set 
free  by  the  action  of  carbonic  acid  from  the  air.  That  no  bleaching 
occurs  after  adding  the  NaOH  settles  this  point,  because  no  free 
HOC1  can  remain  in  presence  of  the  base 

This  experiment  shows  that  the  salt  of  hypochlorous  acid  is 
more  stable  than  the  acid  itself 

2.  Hypobromites.     Add  a  few  drops  of  bromine  to  2  cc. 
of  6-n  NaOH  diluted  with  5  cc.  of  water.     The  red  color  of 
the  bromine  disappears.     Dip  colored  cloth  and  litmus  paper 
in  this  solution  and  then  in  dilute  H2SO4,  to  show  that  it 
bleaches  in  the  same  way  as  a  hypochlorite  solution. 

Add  a  part  of  the  solution  to  2  cc.  of  NH4OH  and  notice 
that  there  is  effervescence,  the  escaping  gas  being  non-com- 
bustible and  a  non-supporter  of  combustion  (nitrogen). 

Add  the  rest  of  the  solution  to  a  solution  of  J  gram  of  urea 
in  5  cc.  of  water  and  notice  a  similar  effervescence. 

Hypobromites  are  very  similar  in  properties  to  hypochlorites 
and  oxidize  organic  coloring  substances.     Ammonia  and  urea  are 
oxidized,  even  in  alkaline  solution,  by  hypobromite, 
3NaOBr  +  2NH3  -» 3NaBr  +  3H20  +  N2, 
3NaOBr  +  CO(NH2)2  -*  3NaBr  +  C02  +  2H2O  +  N2. 

3.  Chlorates  and  Bromates.     The  formation  of  chlorate 
and  bromate  by  heating  hypochlorite  and  hypobromite  respec- 


EXPERIMENTS  231 

tively,  is  illustrated  in  Preps.  36  and  37.  Test  tube  experi- 
ments may  be  tried,  adding  excess  of  chlorine  and  bromine 
respectively  to  hot  6-n  KOH,  boiling  a  little  and  cooling, 
whereupon  the  sparingly  soluble  potassium  chlorate  and 
potassium  bromate  crystallize  out. 

4.  Bromic  and  lodic  Acids.     To  a  globule  of  carbon  di- 
sulphide  in  a  test  tube  add  a  few  drops  of  bromine  water  until, 
after  shaking,  the  globule  has  assumed  a  distinct  red  color. 
Add  5  cc.  water  and  then  chlorine  water,  a  few  drops  at  a  time, 
shaking  after  each  addition.     The  red  color  bleaches  and 
finally  disappears  altogether. 

Repeat  substituting  iodine  for  bromine  and  note  that  the 
deep  violet  color  which  the  free  iodine  imparts  to  the  globule 
of  carbon  disulphide  is  likewise  bleached  on  shaking  the  glob- 
ule with  chlorine  water. 

• 

Recall  Exp.  11,  Chap.  IV,  which  showed  that  chlorine  oxidizes 
bromide  and  iodide  respectively  to  free  bromine  and  iodine.  The 
present  experiment  carries  the  oxidation  still  further  and  raises 
the  valence  of  each  element  to  +  V. 

5C12  +  Br2  +  6H2O  ->  2HBrO3  +  10HC1, 
5C12  +  I2  +  6H2O  -i  2HIO3  +  10HC1. 

Bromic  and  iodic  acids  are  colorless;    they  are  very  soluble  in 
water,  and  highly  ionized. 

5.  Properties    of   Potassium    Chlorate,     (a)    Heat    some 
potassium  chlorate  in  a  test  tube.     It  melts  and  soon  after- 
wards the  melted  salt  effervesces.     A  glowing  splinter  thrust 
into  the  gas  bursts  into  flame. 

(6)  Place  a  very  small  crystal  of  potassium  chlorate  on  a 
watch  glass,  (Danger!  use  of  a  larger  amount  may  cause  a 
violent  explosion),  place  2  drops  of  36-n  H2SO4  on  an  adjacent 
part  of  the  glass  and  let  it  flow  until  it  touches  the  crystal. 
The  latter  dissolves  with  effervescence,  a  deep  yellow  gas 
with  a  very  strong  odor  somewhat  resembling  that  of  chlorine 
being  formed.  Part  of  this  gas  dissolves  in  the  sulphuric  acid 
coloring  it  a  deep  brownish  yellow. 

(c)  Add  6-n  H2SO4  to  a  dilute  potassium  chlorate  solution. 

In  (c)  the  ions  of  chloric  acid  are  brought  together  and  thus  a 
solution  of  chloric  acid  containing  also  the  ions  of  potassium  sul- 


232         OXY-ACIDS  AND   SALTS  OF  THE  NON-METALS 

phate  is  obtained.  Chloric  acid  is  a  soluble,  highly  ionized  acid 
of  a  stability  comparable  to  that  of  nitric  acid. 

In  (6)  chloric  acid  is  formed,  but  the  concentrated  sulphuric  acid 
acts  as  a  dehydrating  agent.  The  oxide  C12O5,  which  is  obviously 
the  anhydride  of  chloric  acid,  however,  does  not  appear;  it  breaks 
down  into  the  oxide  C1O2,  which  is  the  dark  yellow  gas,  and  free 
oxygen.  Chlorine  dioxide  is  not  the  anhydride  of  any  known  acid, 
but  it  dissolves  in  water  and  forms  equivalent  amounts  of  chlorous 
and  chloric  acids,  in  which  the  valence  of  chlorine  is  +  III  and 
+  V  respectively, 

2C102  +  H20  -»  HC102  +  HC103. 

Chlorine  dioxide  is  extremely  explosive  and  it  is  very  dangerous 

to  make  larger  amounts  of  it  than  that  directed  in  this  experiment. 

The  instability  of  the  oxy-compounds  is  further  shown  in  (a) 

in  which  chlorine  is  reduced  from  a  valence  of  +  V  to  one  of  —  I. 

6.  Reduction  of  lodic  Acid,  (a)  Dissolve  a  little  potassium 
iodate  in  water  and  add  some  starch  paste.  No  color  is  ob- 
served. Add  a  little  potassium  iodide  to  part  of  the  mixture, 
and  still  no  color  is  observed.  Now  add  an  acid,  for  example, 
HNO3,  and  observe  that  instantly  the  solution  turns  deep  blue. 

Iodate  and  iodide  ions  alone  have  no  action  on  each  other,  but 
with  hydrogen  ions  present  a  mutual  oxidation  and  reduction  of  the 
iodine  takes  place. 

6H+  +  51-  +  I03-  ->  3H20  +  I2. 

No  oxidation  or  reduction  of  the  hydrogen  occurs,  but  the  hydro- 
gen ion  is  used  up,  which  explains  why  the  presence  of  acid  is 
necessary  to  make  the  reaction  take  place. 

(6)  Dip  filter  paper  in  the  rest  of  the  potassium  iodate- 
starch  mixture  and  suspend  it  in  a  bottle  containing  a  little 
sulphurous  acid.  A  deep  blue  color  immediately  appears  in 
the  paper.  If  the  paper  remains  in  the  bottle  the  blue  is  very 
quickly  bleached,  leaving  the  paper  white  again. 

The  sulphur  dioxide  escaping  from  the  solution  dissolves  in  the 
water  oh  the  paper  and  the  sulphurous  acid  reduces  the  iodide  to 
free  iodine  which  gives  the  blue  color. 

H2SO3  +  2KIO3  -»  K2SO3  +  2HIO3 
2HI03  +  5H2SO3  -»  5H2SO4  +  H2O  +  I2. 


EXPERIMENTS  233 

The  iodine  is  then  reduced  by  more  sulphurous  acid  to  hydriodic 
acid  which  is  colorless, 

Is  +  H20  +  H2S03  ->  H2S04  +  2HI. 

7.  Sulphur  Dioxide.     Burn  some  sulphur  in  a  deflagrating 
spoon  in  a  bottle.     Note  the  characteristic  odor  of  the  gaseous 
product.     Lower  into  the  bottle  a  piece  of  filter  paper  soaked 
in  potassium  iodate-starch  solution  and  note  the  same  effects 
as  in  Exp.  6. 

Sulphur  burns  forming  sulphur  dioxide  SO2.  The  effect  with 
the  potassium  iodate-starch  paper  constitutes  a  convenient  test 
for  sulphur  dioxide. 

8.  Sulphurous  Acid,     (a)  Bubble  sulphur  dioxide  into  a 
bottle  of  water  and  note  that  it  dissolves  freely.     The  solu- 
tion colors  litmus  red  and  it  conducts  electricity  fairly  well. 

(b)  Test  for  Sulphurous  Acid.  To  5  cc.  of  sulphurous  acid 
add  1  cc.  of  6-n  HC1  and  2  cc.  of  BaCl2  solution.  A  small 
precipitate  or  a  slight  clouding  will  probably  occur  at  this 
point,  for  the  sulphurous  acid  solution  invariably  contains 
some  sulphuric  acid  if  it  has  stood  exposed  to  the  air.  Pour 
the  liquid  through  a  filter;  the  nitrate  may  come  through 
cloudy  in  which  case  pour  it  repeatedly  through  the  same 
filter  until  it  is  clear.  To  this  filtrate  add  bromine  and  note 
that  the  red  color  disappears  and  a  white  precipitate  is  formed. 

Sulphurous  acid  is  a  rather  weak  acid  and  presence  of  the  strong 
acid  HC1  prevents  the  formation  of  SO3  ions  in  sufficient  con- 
centration to  precipitate  with  Ba++  ions.  The  first  precipitate 
removes  all  of  the  sulphuric  acid.  Then  the  addition  of  bromine 
oxidizes  the  sulphurous  acid  to  sulphuric  acid  and  another  pre- 
cipitate of  barium  sulphate  is  produced. 

H2S03  +  Br2  +  H2O  -» H2S04  +  2HBr. 

9.  Reducing  Action   of   Sulphurous  Acid.     To   5   cc.   of 

potassium  permanganate  solution  add  sulphurous  acid.     The 
deep  red  solution  becomes  colorless. 

5H2SO3  +  2KMnO4  -*  K2SO4  +  2MnSO4  +  2H2S04  +  3H2O. 

10.  Oxidizing  Action  of  Sulphur  Dioxide  and  Sulphurous 
Acid,     (a)  Fill  a  250  cc.  wide-mouthed  bottle  with  sulphur 


234         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

dioxide  and  lower  into  the  gas  a  burning  strip  of  magnesium 
ribbon  held  by  pincers.  The  magnesium  continues  to  burn 
brilliantly,  forming  the  same  white  smoke  as  if  it  burned  in 
air.  There  is  also  noticed  a  yellowish  deposit  on  the  glass 
(sulphur). 

(6)  Pass  hydrogen  sulphide  into  a  solution  of  sulphurous 
acid.     A  milky  precipitate  is  formed. 

Although  sulphur  has  considerable  affinity  for  oxygen,  its 
strength  in  this  respect  is  far  less  than  that  of  the  active  metal 
magnesium. 

In  (6)  we  have  another  case  similar  to  that  in  Exp.  6  (a)  in  which 
an  element  existing  in  different  states  of  oxidation  mutually 
oxidizes  and  reduces  itself. 

H2SO3  +  2H2S  -» 3H20  +  3S  j 
IV  +  2(-  II) ->0 

11.  Dehydrating  Action  of  Sulphuric  Acid.     Add  10  cc.  of 
36-n  H2SO4  to  5  grams  of  sugar  in  a  porcelain  dish.     If 
necessary  warm  the  mixture  a  little  to  start  a  reaction.     Once 
started  the  reaction  proceeds  vigorously  giving  off  a  good  deal 
of  heat,  and  the  sugar  swells  up  and  grows  black.     Finally, 
there  is  left  a  bulky  brittle  charcoal  like  mass. 

The  formula  of  sugar  Ci2H22On  shows  that  it  contains  hydrogen 
and  oxygen  in  the  proportion  to  form  water.  The  affinity  of  sul- 
phuric acid  for  water  is  so  great  that  it  causes  the  sugar  to  decom- 
pose so  as  to  yield  carbon  anpl  water. 

12.  Oxidizing  Action  of  Sulphuric  Acid.     In  separate  test 
tubes  heat  sulphur,  charcoal,  and  copper  turnings  with  36-n 
H2SC>4,  and  hold  potassium  iodate-starch  paper  in  the  mouth 
of  each  tube.     In  each  case  the  test  paper  is  turned  blue. 

Concentrated  sulphuric  acid  is  an  oxidizing  agent  (cf.  Exps.  13 
and  14  in  Chap.  IV)  it  being  reduced  usually  to  sulphurous  acid 
(S02). 

2H2SO4  +  S  ->  3SO2  +  2H2O 
2H2S04  +  C  ->  2S02  +  C02  +  2H20 
|  H2SO4  +  Cu  -»  CuO  +  S02  +  H2O 
\    CuO  +  H2SO4  ->  CuS04  +  H2O 


EXPERIMENTS  235 

This  experiment  should  be  compared  with  Exps.  6  (6),  8  (6)  and  9,  in 
which  sulphurous  acid  reduces  iodic  acid,  bromine,  and  perman- 
ganate it  itself  being  oxidized  to  sulphuric  acid.  The  latter  is 
not  an  oxidizing  agent  in  dilute  solution. 

13.  Nitric  Acid  as  an  Oxidizing  Agent,  (a)  To  3  grams 
copper  turnings  in  a  test  tube  add  5  cc.  water  and  5  cc.  of  6-n 
HNOs.  Fit  stopper  with  delivery  tube  leading  to  a  trough  of 
water.  Heat  the  test  tube  a  very  little  to  start  the  reaction; 
let  the  gas  go  to  waste  until  it  appears  colorless  in  the  test  tube, 
then  collect  a  test  tube  full  of  it  at  the  water  trough.  Note 
that  the  gas  is  colorless  and  that  it  is  not  soluble  in  the  water. 
Remove  the  test  tube  from  the  trough,  turn  it  mouth  up  and 
hold  a  sheet  of  white  paper  behind  it ;  a  red  gas  is  seen  where 
the  gas  in  the  tube  meets  the  air.  Note  the  suffocating  odor 
of  this  gas. 

Non-oxidizing  acids  have  no  effect  on  copper  but  they  dissolve 
copper  oxide.  Nitric  acid  first  oxidizes  copper,  it  itself  being  re- 
duced to  nitric  oxide,  3Cu  +  2HNO3  ->  3CuO  -f  H2O  +  2NO; 
before  any  further  oxidation  of  the  copper  occurs,  the  nitric  acid 
acts  as  an  acid  with  the  copper  oxide  giving  the  soluble  copper 
nitrate,  and  it  is  for  this  reason  that  we  see  no  direct  evidence  of 
the  intermediate  reaction.  Nitric  oxide  has  the  most  remarkable 
property  of  combining  spontaneously  with  oxygen  at  ordinary 
temperature,  forming  the  deep  reddish  brown  gas  nitrogen  dioxide, 
NO  +  i02  -» N02. 

(6)  To  2  grams  of  zinc  turnings  in  a  test  tube  add  20  cc.  of 
water  and  2  cc.  of  6-n  HNO3.  Warm  the  mixture  but  not 
to  boiling;  then  let  it  stand  5  minutes.  Test  the  solution  by 
pouring  it  from  the  undissolved  zinc,  adding  sodium  hydroxide 
in  excess,  and  warming,  when  the  odor  of  ammonia  becomes 
apparent. 

Zinc  is  a  much  more  powerful  reducing  agent  than  copper; 
furthermore,  zinc  displaces  from  an  acid,  hydrogen,  which,  at  the 
moment  of  its  displacement  is  in  the  atomic  or  "  nascent  "  condi- 
tion, and  especially  active.  Under  these  conditions  the  nitrogen 
is  reduced  to  its  lowest  limit  represented  by  ammonia.  The 
ammonia  does  not  escape  from  the  solution  because  it  combines 
with  the  excess  of  nitric  acid. 


236         OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

4Zn  +  8HNO3  -» 4Zn(NO3)2  +  8H 
HN03  +  8H  ->  3H20  +  NH3 

NH3  +  HNO3  ->  NH4NO3 

4Zn  +  10HNO3  -»  4Zn(NO3)2  +  3H2O  +  NH4NO3 

4Zn  0  to  -ffh        4  X  (+  2)  =  +  8 

IN        +  V  to  -  III =_-  8 

Total  change  =       0 

Explain  how  the  final  step  in  the  experiment  is  an  example  of  the 
displacement  of  a  weak  base  from  its  salt  by  a  stronger  base  and 
that  it  involves  no  change  in  the  primary  valence  of  any  of  the 
elements. 

14.  Nitrous  Acid,  (a)  Dissolve  J  gram  of  sodium  nitrite  in 
5  cc.  of  ice  water  and  add  1  cc.  of  cold  6-n  H2S04.  A  blue 
solution  results  which  at  0°  effervesces  very  slowly.  Add  a 
few  drops  of  this  solution  to  a  half  test  tube  of  water  con- 
taining a  few  drops  of  iodide-starch  solution.  An  intense 
blue  color  is  produced.  Add  a  few  drops  of  sodium  nitrite 
solution  alone,  of  sulphuric  acid  alone,  and  of  nitric  acid  alone 
to  tubes  made  up  with  a  similar  amount  of  iodide-starch  and 
note  that  in  none  of  these  cases  is  any  color  produced.  (To 
produce  no  effect  the  nitric  acid  must  be  free  from  nitrous 
acid  and  it  should  be  taken  from  a  special  bottle  prepared  for 
this  experiment.) 

Allow  the  rest  of  the  potassium  nitrite-sulphuric  acid  mix- 
ture to  warm  up  to  room  temperature.  Note  that  it  effer- 
vesces rather  strongly,  giving  off  a  red  gas  and  that  the  blue 
color  quickly  disappears. 

Nitrous  acid  is  of  about  the  same  strength  as  acetic  acid  and  it 
is  blue  in  color.  It  is  formed  when  its  ions  are  brought  together. 
It  is  very  unstable,  decomposing  mainly  according  to  the  equation, 

3HNO2  -»  HNO3  +  2NO  +  H2O, 

the  NO  giving  the  red  color  when  it  comes  in  contact  with  air. 
Nitrous  acid  is  an  oxidizing  agent,  liberating  iodine  from  hydriodic 
acid, 

2HN02  +  2HI  -> 2H20  +  2NO  +  I2. 

Since  nitric  acid  does  not  liberate  iodine  under  similar  circum- 
stances, the  nitrous  acid  is  shown  to  be  the  more  vigorous  oxidizing 


GENERAL  QUESTIONS  VIII  237 

agent.  This  is  quite  in  accord  with  what  we  have  already  found 
in  other  cases,  namely,  that  the  stability  of  the  acid  is  greatest  the 
higher  the  valence  of  the  principal  element.  It  is  also  seen  that 
the  nitrite  ion  alone  is  not  the  oxidizing  agent,  the  presence  of 
hydrogen  ions  is  necessary. 

(6)  Reducing  Action  of  Nitrous  Acid.  To  about  0.1  gram 
sodium  nitrate  and  100  cc.  of  cold  water  in  a  beaker  add  10  cc. 
of  6-n  H2S04.  To  this  solution  add  drop  by  drop  with  con- 
stant stirring  0.1  formal  KMnO4,  until  the  pink  color  pro- 
duced by  each  drop  disappears  more  and  more  slowly  and 
finally  1  drop  more  produces  a  permanent  pink  color. 

The  permanganate  oxidizes  the  nitrous  acid  to  nitric  acid, 

5HNO2  +  2KMnO4  +  3H2SO4  ->  K2S04  +  2MnS04  + 
3H20 


The  reaction  proceeds  so  sharply  to  completion  and  the  intense 
color  of  the  permanganate  serves  as  such  an  excellent  indicator, 
that  it  is  very  easy  to  estimate  the  amount  of  nitrite  from  the 
volume  of  a  solution  of  KMn04  of  known  strength  that  it  will 
decolorize. 

GENERAL  QUESTIONS  VIII 
THE  OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

1.  Make  a  table,  in  the  first  column  of  which  place  the  formulas 
of  hypochlorous  acid,  chlorous  acid,  chloric  acid,  perchloric  acid, 
the  four  corresponding  acids  of  bromine  and  of  iodine  with  two 
per-iodic  acids  differing  in  degree  of  hydration,  sulphurous  acid, 
sulphuric  acid,  nitrous  acid,  nitric  acid.     In  the  second  column 
give  the  valence  of  the  element  in  the  particular  acid;  in  the  third 
column  the  formula  of  the  anhydride;  in  the  fifth  column  the  word 
gas,  liquid,  solid,  or  hyp.,  according  as  to  the  state  of  aggregation 
of  the  anhydride  at  ordinary  temperature,  hyp.  signifying  hypo- 
thetical or  non-existent;  in  the  sixth  column  the  percent  ionization 
of  the  acid  in  0.1  normal  solution,  —  if  no  exact  figure  can  be 
found  specify  whether  very  weak,  weak,  or  strong. 

2.  Give  the  names  of  the  oxides  having  the  formulas  N20,  NO, 
NO2,  C102.     Give  a  brief  account  of  the  properties  of  each,  includ- 
ing state  of  aggregation,  stability,  ability  to  support  combustion, 
behavior  with  water. 


238        OXY-ACIDS  AND  SALTS  OF  THE  NON-METALS 

3.  When  an  element  forms  a  series  of  oxy-acids,  what  rule  seems 
to  hold  connecting  the  valence  with  the  acid  strength? 

4.  Compare  the  affinity  of  sulphur  dioxide  and  sulphur  trioxide 
for  water,  giving  whatever  data  you  can  find.     Make  a  similar 
comparison  of  C12O  and  C1207,  N203  and  N2O5.     Can  any  general 
rule  be  stated  to  cover  these  cases? 


CHAPTER  IX 

ELEMENTS   OF  THE   GROUP   IV  OF  THE 
PERIODIC   SYSTEM 

This  group  stands  in  the  middle  of  the  Periodic  Table  of  the 
elements,  and  in  it  the  difference  in  properties  between  the  elements 
of  Family  A  and  Family  B  is  at  a  minimum.  As  in  the  case  of 
Group  III,  therefore,  the  whole  group  is  considered  under  one 
heading. 

The  elements  of  this  group  which  come  most  to  our  attention 
in  everyday  life  are  carbon,  silicon,  tin,  and  lead.  Carbon  and 
silicon  are  the  first  two  members  and  are  exclusively  acid-forming 
elements,  although  the  acids  formed  are  not  strong  ones.  Tin 
and  lead  are  the  last. two  members  of  Family  B  and  are  in  the 
main  base-forming;  they  are  comparable  in  this  respect  with  the 
heavy  metals  already  considered  under  Groups  I  and  II.  In  this 
group  the  elements  of  low  atomic  weight  are  exclusively  acid- 
forming,  while  the  elements  of  high  atomic  weight  are  almost 
entirely  base-forming.  Between  these  extremes  there  is  an  almost 
regular  gradation  of  properties. 

Titanium,  although  it  is  a  fairly  abundant  constituent  of  the 
earth's  crust,  is  an  element  of  comparatively  little  importance. 

Cerium  and  thorium,  the  heaviest  two  elements  of  Family  A, 
have  acquired  some  importance  on  account  of  the  use  of  their 
oxides  in  incandescent  gas-lighting  mantles. 

PREPARATION  42 
PRECIPITATED  SILICA 

Carbon  dioxide  and  silicon  dioxide  are  chemically  very  similar 
to  each  other  in  that  both  form  weak  acids,  that  of  silicon,  the 
heavier  element,  naturally  being  the  weaker  acid.  On  the  other 
hand  these  oxides  are  very  dissimilar  in  their  physical  properties, 
one  being  a  gas,  and  the  other  a  solid  with  an  extremely  high 
melting  point. 

The  mineral  quartz  is  crystallized  silicon  dioxide.  Sea  sand 

239 


240  ELEMENTS  OF  GROUP  IV 

consists  mostly  of  rounded  grains  of  broken  quartz.  If  finely 
ground  quartz  or  sand  is  fused  for  a  long  time  with  sodium  car- 
bonate, the  weaker,  but  non-volatile,  acid  anhydride  displaces 
the  carbon  dioxide,  and  sodium  silicate  is  obtained.  This  is  a 
glass-like  substance,  which,  however,  can  slowly  be  dissolved  by 
water  heated  under  pressure.  The  solution  so  obtained  is  evapo- 
rated to  a  syrup  like  consistency  and  is  sold  on  the  market  under 
the  name  of  "  water  glass."  Several  grades  of  water  glass  of  differ- 
ent ratios  Na^O:  SiO2  are  sold,  but  perhaps  the  most  common 
grade  approximates  the  composition  Na^Si^  (=  Na20.4Si02  = 
Na2Si03.3Si02). 

The  addition  of  an  acid  to  a  sodium  silicate  solution  causes  a 
separation  of  silicic  acid  which  appears  as  a  jelly-like  substance. 
Ortho-silicic  acid  has  the  composition  H4Si04,  meta-silicic  acid, 
H2Si03j  the  acid  corresponding  to  the  sodium  salt  of  the  above 
formula,  H^Si^g.  Suspended  in  water  these  different  silicic  acids 
are  more  or  less  easily  interchangeable  one  into  the  other,  but, 
if  silicic  acid  is  heated,  it  loses  all  its  water,  and  becomes  the  an- 
hydride. The  anhydride  practically  will  not  take  on  water  again 
to  form  acids. 

The  very  finely  divided  anhydride  prepared  by  precipitating 
and  drying  silicic  acid  is  more  reactive  than  the  most  finely 
powdered  quartz,  and  it  finds  use  as  a  reagent  in  certain  analytical 
tests. 

Materials:     water-glass,  25  cc. 

6-n  HC1. 
Apparatus:  6-inch  evaporating  dish. 

tall  600  cc.  beaker. 

suction  filter. 

Procedure:  To  25  cc.  of  water  glass  in  a  porcelain  dish  add 
25  cc.  of  water  and  slowly  stir  in  6-n  HC1  until  the  acid  is  in  excess. 
The  liquid  first  coagulates  to  a  jelly,  then  the  jelly  hardens  and 
breaks  up  on  stirring  into  seemingly  dry  lumps,  and  later  after 
an  excess  of  acid  is  added  (about  30  cc.  in  all)  the  mass  grows 
partially  fluid  again.  Place  the  dish  on  a  water  bath  or  a  hot 
plate  to  evaporate  to  complete  dry  ness.  If  a  hot  closet  at  130° 
is  available  the  dish  and  contents  should  be  baked  for  one  hour 
at  this  temperature.  Otherwise  heat  the  dish  over  a  flame  for 
fifteen  minutes  avoiding,  if  possible,  letting  any  part  of  the  con- 


STANNOUS  CHLORIDE  241 

tents  get  above  150°  as  this  would  render  traces  of  iron  oxide 
very  difficult  to  redissolve  in  acid.  While  the  dish  is  still  warm 
moisten  the  contents  completely  with  12-n  HC1.  Let  it  digest 
for  15  minutes.  Wash  the  contents  of  the  dish  into  a  tall  beaker 
and  let  the  silica  settle.  Wash  by  decantation  several  times  and 
collect  the  silica  on  the  suction  filter,  washing  it  well  on  the  filter. 
Dry  the  product  and  put  it  up  in  a  cork-stoppered  test  tube. 

QUESTIONS 

1.  Quartz  can  be  melted  like  glass,  but  at  a  much  higher 
temperature,  and  many  kinds  of  chemical  apparatus  are  made 
of  fused  quartz.    .Dishes  made  of  fused  quartz  are  used  for  boiling 
concentrated  sulphuric  acid.     Why  cannot  they  be  used  equally 
well  for  concentrating  caustic  alkalies? 

2.  Mix  \  gram  of  the  precipitated  silica  with  1  gram  of  powdered 
calcium  fluoride.     Place  the  mixture  in  a  test  tube,  moisten  it  with 
36-n  H2SO4  and  warm  it  gently  under  the  hood.     Dip  a  stirring 
rod  in  water  and  lower  it,  with  a  drop  adhering,  into  the  gas 
in  the  test  tube.     Note  the  precipitate  that  forms  in  the  drop  of 
water.     Explain  all  of  the  reactions,  with  equations,  and  state 
what  rather  unusual  properties  are  shown  by  this  experiment  to 
be  possessed  by  hydrofluoric  acid  and  by  silicon  tetrafluoride. 

PREPARATION  43 
STANNOUS  CHLORIDE  (SnCl2.2H2O) 

This  salt  can  be  prepared  by  the  action  of  hydrochloric  acid 
upon  metallic  tin,  but  since  the  action  is  exceedingly  slow,  it  is 
hastened  by  the  addition  of  a  very  small  quantity  of  nitric  acid, 
which  oxidizes  the  tin.  Nitric  acid  is  ordinarily,  by  its  action 
upon  a  metal,  reduced  only  to  the  oxide  NO ;  but  in  the  course  of 
this  preparation  no  red  fumes  of  oxides  of  nitrogen  are  found  to 
escape,  because,  under  the  influence  of  tin  and  stannous  chloride, 
the  reduction  does  not  stop  at  nitric  oxide,  but  continues  to  the 
lowest  possible  step,  which  is  ammonia  or  in  this  case  its  salt, 
ammonium  chloride.  Stannous  salts  are  oxidized  quite  readily 
to  stannic  by  the  oxygen  of  the  air;  to  prevent  this  happening 
during  the  evaporation  of  the  solution,  an  excess  of  metallic  tin  is 
kept  in  the  liquid. 


242  ELEMENTS  OF  GROUP  IV 

Materials:     feathered  tin,  100  grams. 

12-n  HC1,  175  cc. 

6-n  HNO3,  25  cc. 

500  cc.  casserole. 
Apparatus:  shredded  asbestgs  for  filter. 

suction  filter. 

6-inch  evaporating  dish. 

Procedure:  Place  100  grams  of  feathered  tin  in  a  large  casserole, 
cover  with  175  cc.  of  12-n  HC1  and  add  (at  the  hood)  25  cc.  of  6-n 
HNO3,  a  little  at  a  time,  during  a  period  of  10  minutes.  Then 
concentrate  the  solution,  by  boiling  over  a  free  flame,  to  a  volume 
of  90-100  cc.,  at  which  point  a  crystal  scum  will  form  on  blowing 
on  the  surface  of  the  hot  liquid.  There  should  still  be  left  a  small 
amount  of  undissolved  metal.  If  at  any  time  during  the  evapo- 
ration all  the  tin  becomes  used  up,  add  a  little  more.  Prepare  an 
asbestos  filter  (Note  4  (d),  page  8),  moisten  it  with  concentrated 
hydrochloric  acid,  and  filter  the  concentrated  stannous  chloride 
solution  before  it  has  cooled  to  below  60-70°.  Finally,  rinse  out 
the  casserole  with  15  cc.  of  concentrated  hydrochloric  acid  and 
pour  this  liquid  through  the  filter,  letting  it  mix  with  the  main 
part  of  the  solution.  If  during  the  filtration  the  liquid  stops  flow- 
ing, due  to  crystals  separating  in  the  filter,  add  5-10  cc.  of  boiling 
water.  Pour  the  solution  into  a  6-inch  evaporating  dish,  and 
leave  it  to  evaporate  slowly  at  room  temperature  in  a  place  ex- 
posed to  the  air  but  protected  from  dust.  (The  solubility  of 
stannous  chloride  decreases  very  rapidly  with  decreasing  temper- 
ature. Hence  it  is  advantageous  to  carry  out  the  crystallization 
in  as  cool  a  place  as  possible.)  When  a  good  crop  of  crystals  has 
formed,  pour  off  the  liquid  into  another  dish,  spread  the  crystals 
on  paper  towels  and  allow  them  to  dry.  It  is  to  be  remembered 
that  stannous  chloride  is  extremely  soluble  in  water  and  that  the 
composition  of  the  mother  liquor  is  not  far  different  from  that  of 
the  crystals  of  SnCl2.2H2O  which  separate.  Heat  the  remaining 
solution  carefully  just  to  the  boiling  temperature,  but  do  not  allow 
it  to  boil  more  than  a  moment.  In  this  way  sufficient  water  and 
hydrochloric  acid  are  expelled  to  allow  another  crop  of  crystals 
to  form.  If  too  much  hydrochloric  acid  is  expelled  by  the  evap- 
oration and  an  indistinctly  crystalline  precipitate  of  basic  salt 
separates  on  cooling,  add  a  few  drops  of  hydrochloric  acid  and 


STANNOUS  CHLORIDE  243 

redissolve  the  salt  by  warming.  Set  the  solution  aside  to  cool  and 
evaporate,  as  before,  and  collect  another  crop  of  crystals.  By 
repeating  this  process  once  or  twice  more,  almost  the  entire  mother 
liquor  should  be  used  up  and  nearly  the  calculated  yield  of  stannous 
chloride  should  be  obtained. 

QUESTIONS 

1.  Explain  why  during  this  preparation  no  red  oxides  of  nitrogen 
are  seen  to  escape  in  consequence  of  the  reduction  of  nitric  acid 
by  the  metal.     If  nitric  acid  is  reduced  to  NH3,  show  how  many 
more  equivalents  of  oxygen  it  will  yield  for  the  oxidation  of  the 
tin  than  if  it  were  reduced  only  to  NO. 

To  test  for  the  presence  of  ammonium  salt  in  the  product,  take 
about  1  gram  of  the  crystals;  dissolve  in  10  cc.  of  water  in  a  small 
beaker.  Add  sodium  hydroxide  solution  until  the  precipitate 
first  formed  redissolves.  Place  over  the  beaker  a  watch  glass,  on 
the  under  side  of  which  is  stuck  a  piece  of  moistened  red  litmus 
paper.  Place  some  cold  water  in  the  hollow  of  the  watch  glass, 
and  warm  the  solution  in  the  beaker  very  gently.  What  observa- 
tion will  indicate  the  presence  of  ammonium  salt,  and  why? 

2.  Dissolve  1  gram  of  stannous  chloride  crystals  in  1  to  2  cc.  of 
cold  water.     Then  add  a  considerable  amount  of  water.     What  is 
the  precipitate?     What  can  be  added  to  prevent  its  formation? 

3.  To  a  cold  solution  of  stannous  chloride  add  sodium  hydroxide 
until  it  has  redissolved  the  precipitate  first  formed.     Write  the 
equation.     Save  the  solution. 

4.  Pour  the  solution  saved  from  Experiment  3  over  a  little 
bismuth  hydroxide  on  a  filter  paper.     (The  latter  can  be  precipi- 
tated for  the  occasion.)     Compare  the  action  with  that  of  stannous 
chloride  on  mercuric  chloride. 

5.  Prepare  a  very  concentrated  cold  solution  of  sodium  stan- 
nite:  Dissolve  1  gram  of  stannous  chloride  in  1  cc.  of  water.     Dis- 
solve a  small  lump  of  sodium  hydroxide  in  its  own  weight  of  water, 
and  add  this  solution,  a  drop  at  a  time,  to  the  first  solution  — 
cooling  all  the  while  under  the  water  tap  —  until  the  precipitate 
at  first  formed  redissolves.     Then  heat  the  solution.     Compare 
the  action  with  that  in  Experiment  4. 

6.  In  preparing  a  solution  of  stannous  chloride  for  a  laboratory 
reagent,  what  is  the  necessity  of  adding  hydrochloric  acid  and  of 
placing  a  piece  of  metallic  tin  in  the  bottle? 


244  ELEMENTS  OF  GROUP  IV 

PREPARATION  44 
STANNIC  SULPHIDE  (MOSAIC  GOLD) 

Stannic  sulphide,  Sn$2,  is  the  higher  sulphide  of  tin,  and  can 
be  prepared  by  direct  combination  of  the  metal  or,  still  better,  of 
the  lower  sulphide,  SnS,  with  sulphur.  Under  ordinary  conditions 
these  two  substances  will  not  react  at  a  temperature  below  that 
which  will  decompose  stannic  sulphide.  If,  however,  they  are 
mixed  with  ammonium  chloride  the  presence  of  this  substance 
makes  possible  the  combination  at  a  lower  temperature.  The 
stannic  sulphide  formed  in  this  way  appears  as  soft,  glistening, 
yellow  crystals.  It  is  used  as  a  bronzing  powder,  and  is  known 
under  the  name  of  mosaic  gold.  In  physical  properties  it  is  very 
different  from  the  stannic  sulphide  which  can  be  precipitated  by 
hydrogen  sulphide  from  a  solution  of  stannic  chloride. 

Stannous  chloride,  the  raw  material,  if  it  is  fresh,  is  completely 
soluble  in  a  very  little  water;  with  much  water  it  hydrolyzes 
somewhat  with  precipitation  of  basic  stannous  chloride,  Sn(OH)Cl. 
If  the  stannous  chloride  is  old,  it  has  probably  become  partially 
oxidized  to  stannic  chloride,  and  the  latter  extensively  hydrolyzed 
to  insoluble  stannic  acid  H2SnO3.  A  clear  solution  is  therefore  not 
obtained  when  the  stannous  chloride  is  treated  with  a  large  amount 
of  water.  Nevertheless  the  addition  of  ammonium  sulphide  con- 
verts all  of  the  tin  to  sulphide,  the  sulphides  being  very  much  less 
soluble  than  the  products  of  the  hydrolysis. 

Materials:     stannous  chloride,  SnCl2.2H20,  45  grams  =  |  F.W. 

hydrogen  sulphide  (Note  13,  page  19). 

powdered  sulphur. 

ammonium  chloride. 
Apparatus:  large  common  bottle. 

hydrogen  sulphide  generator. 

Procedure:  Stannous  Sulphide.  Place  the  stannous  chloride  in 
a  beaker  and  treat  it  with  its  own  weight  of  water.  If  it  dissolves 
completely  pour  it  into  a  large  common  bottle  and  dilute  it  with 
1500  cc.  of  hot  water.  If  it  does  not  all  dissolve  decant  the  solu- 
tion into  the  bottle,  crush  any  hard  lumps  with  a  pestle,  and  treat 
the  residue  successively  with  small  amounts  of  water  pouring  the 
easily  floatable  suspension  each  time  into  the  bottle.  Make  up 


STANNIC  SULPHIDE 


245 


the  volume  to  1500  cc.  and,  under  the  hood,  pass  hydrogen  sul- 
phide in  through  a  delivery  tube  leading  to  the  bottom  of  the  bottle. 
Continue  this  treatment  until  the  solution  is  saturated  with  the 
gas,  when,  after  vigorous  shaking,  it  will  smell  of  hydrogen  sul- 
phide. Let  the  precipitate  settle,  decant  off  the  clear  solution 
and  throw  the  sludge  on  to  a  gravity  filter.  Do  not  wash  the 
sludge,  merely  let  it  drain  thoroughly.  Without  tearing  the  filter, 
remove  it  from  funnel  and  spread  it  out  on  paper  towels  on  the 
hot  plate.  Pulverize  the  product  when  it  is  dry. 

Stannic  Sulphide:  (Save  2  grams  of  stannous  sulphide  for  an 
experiment).  Grind  the  rest  together  with  one  half  its  weight 
of  sulphur  and  f  its  weight  of  ammonium  chloride.  Bring  the 
mixture  into  an  8-inch  test  tube.  Close  the  tube  with  a  stopper 
bearing  a  short  piece  of  glass  tube  drawn  out  to  a  capillary  and 
bend  downward  at  right  angles.  The  capillary  is  to  relieve  any 
pressure  caused  by  heating  and  at  the  same  time  prevent  outside 
air  entering  the  tube  during  the  heating.  Lay  the  tube  in  a  sand 
bath  pan  containing  a  J-inch  layer  of  sand.  Then  heap  sand  over 
the  part  of  the  tube  containing  the  charge.  Heat  the  sand  bath, 
first  rather  moderately  for  15  minutes,  then  for  1  hour  so  that  the 
bottom  of  the  iron  pan  is  bright  red.  Cool,  break  the  tube,  sepa- 
rate the  layer  of  stannic  sulphide  from  the  dirty  colored  material 
on  top,  pulverize  the  product  and  put  it  up  in  a  2-ounce  cork- 
stoppered  bottle. 

QUESTIONS 

1.  Experiment:  Treat  J  gram  powdered  stannous  sulphide  with 
sodium  sulphide  (Na2S)  solution  warming  for  about  3  minutes. 
Does  the  solid  dissolve?     Then  add  about  -^  gram  of  powdered 
sulphur  and  warm  a  little  longer.     Does  the  stannous  sulphide 
now   go   into   solution?     Write  equations  and  explain  how  the 
sulphur  could  have  caused  the  stannous  sulphide  to  dissolve. 

Finally  acidify  the  solution.  What  is  the  precipitate?  Equa- 
tion? 

Test  portions  of  this  precipitate  to  see  if  it  will  dissolve  in 
6-normal  HC1,  —  in  warm  (NH4)2S.  Does  the  mosaic  gold  dis- 
solve in  these  reagents?  Can  you  explain  the  difference? 

2.  Write  ionized  equations  for  (a)  the  precipitation  of  basic 
stannous  chloride  SnOHCl,  when  a  solution  of  SnCl2  is  diluted 
with  a  large  amount  of  water.     (6)  the  conversion  of  SnOHCl  to 
SnS  when  this  precipitate  is  treated  with  (NH^S  solution. 


246  ELEMENTS  OF  GROUP  IV 

PREPARATION  45 
STANNIC  CHLORIDE  (ANHYDROUS) 

Anhydrous  stannic  chloride,  SnCl4,  is  prepared  by  the  action 
of  dry  chlorine  gas  upon  metallic  tin.  It  is  a  colorless,  very 
mobile  liquid  which  boils  at  114°.  At  ordinary  temperature  it 
has  a  considerable  vapor  pressure  and  the  vapor,  reacting  strongly 
with  the  water  vapor  of  the  air,  gives  rise  to  dense  fumes.  With 
liquid  water  stannic  chloride  reacts  violently.  If  a  limited  amount 
of  water  is  added  with  caution  it  is  possible  to  obtain  solid  hydrates 
of  the  composition  SnCl4.3H2O,  or  SnCl4.5H2O.  These  hydrates 
dissolve  to  form  apparently  clear  solutions  but  the  salt  is  very 
extensively  hydrolyzed,  the  rather  complex  hydrolysis  products 
remaining  for  the  most  part  in  a  colloidal  condition. 

Materials:    feathered  tin,  150  grams. 

chlorine. 

tin  foil. 
Apparatus:  350  cc.  tubulated  retort. 

5  300  cc.  bottles. 

chlorine  generator  (1500  cc.  flask). 

36-inch  condenser. 

200  cc.  distilling  flask. 

150  cc.  container  for  stannic  chloride:  the  neck 
must  be  previously  drawn  out  to  a  narrow  tube 
that  can  be  quickly  sealed  off  in  blast  flame. 

delivery  tubes,  connectors  and  rubber  stoppers  as 
in  diagram. 

This  preparation  is  only  to  be  attempted  if  four  consecutive 
hours  are  available  in  the  laboratory,  and  even  then,  the  apparatus 
should  be  assembled  at  a  previous  exercise. 

Fit  up  a  chlorine  generator  with  a  1500  cc.  flask  in  which  place 
33  percent  in  excess  of  the  calculated  quantity  of  manganese 
dioxide.  The  gas  is  to  be  passed  through,  first  one  wash  bottle 
containing  water,  and  then  two  wash  bottles  containing  concen- 
trated sulphuric  acid.  Place  150  grams  of  tin  in  a  tubulated  retort, 
and  place  the  retort  on  a  sand  bath.  The  neck  of  the  retort 
should  pass  into  a  long  condenser,  and  the  latter  should  empty 
into  a  200  cc.  distilling  flask,  in  which  has  been  placed  some  tin 


STANNIC  CHLORIDE 


247 


foil.  Connect  the  side  arm  of  the  flask  with  a  bottle  containing 
sodium  hydroxide  solution  to  absorb  the  waste  chlorine.  The  tube 
entering  the  bottle  should  not  dip  into  the  liquid,  but  should  reach 
down  to  near  its  surface;  a  safety  tube  should  be  supplied,  and 
the  exit  tube  should  dip  into  a  solution  of  sodium  hydroxide  in 


Chlorine    Water 
Generator 


H2SOt 


&-N  NaOH 


FIG.  20 


another  bottle.  Into  the  retort  should  be  fitted  the  tube  supply- 
ing chlorine  from  the  generator  and  wash  bottles,  and  this  should 
reach  nearly  to  the  center  of  the  surface  of  the  tin,  which  is  to  be 
melted  before  the  action  is  started.  Glass  tubing  is  to  be  used 
throughout,  and  where  connections  are  made  with  rubber  the 
ends  of  the  glass  tubes  should  be  brought  close  together.  Before 
beginning  to  generate  the  chlorine,  the  whole  apparatus  must  be 
proved  to  be  tight,  so  that  none  of  this  gas  can  escape  into  the 
laboratory. 

Procedure:  Melt  the  tin.  Commence  the  generation  of  chlorine 
and  regulate  it  so  that  the  tin  in  the  retort  can  be  seen  to  burn 
quietly.  Continue  the  action  until  all  the  tin  has  disappeared 
and  the  tin  tetrachloride  has  been  caught  in  the  receiving  flask. 
Remove  the  neck  of  the  retort  from  the  condenser,  and  insert 
instead  a  stopper  with  a  tube  leading  to  the  bottles  already  used 
for  absorbing  waste  chlorine.  Close  the  side  arm  of  the  receiving 
flask  and,  with  the  condenser  still  in  the  same  position,  boil  the  tin 
tetrachloride  until  it  is  colorless  (it  contains  a  large  amount  of 
dissolved  chlorine  and,  on  boiling,  this  reacts  with  the  tin  foil). 
Change  the  position  of  the  flask  and  condenser,  and  distill  the  tin 
tetrachloride  into  the  prepared  container.  During  the  distillation 
this  container  should  not  be  open  to  the  air,  but  should  be  con- 


248 


ELEMENTS  OF  GROUP  IV 


nected  by  a  tube  to  the  absorbing  bottles  already  used.  When 
the  liquid  is  all  distilled,  seal  the  neck  of  the  container  at  a  blast 
lamp,  so  that  the  preparation  can  be  preserved  out  of  contact 
with  the  air. 

& 

QUESTIONS 

1.  What  is  the  purpose  of  the  wash  bottles  as  arranged  for  the 
chlorine  gas? 

2.  Suggest  how  hydrated  stannic  chloride  SnCl4.5H2O  might 
be  made,  starting  with  stannous  chloride. 

3.  What  happens  if  hydrated  stannic  chloride  is  heated?     (Com- 
pare aluminum  chloride  and  bromide,  Preps.  26  and  27.) 

PREPARATION  46 
LEAD  NITRATE 

Lead  nitrate  is  one  of  the  most  readily  prepared  salts  of  lead, 
since  it  is  of  moderate  solubility  and  can  be  obtained  in  well- 
formed  anhydrous  crystals,  Pb(NO3)2.  In  it  lead  appears  in  its 
usual  state  of  oxidation,  which  corresponds  to  that  of  the  oxide 
PbO;  indeed,  the  salt  is  actually  prepared  by  treating  this  oxide 
(litharge)  with  nitric  acid. 


A  saturated  solution  contains  for  each  100  grams  of  water  the  given 
number  of  grams  of  lead  nitrate 


Temperature  

0° 

10° 

18° 

25° 

50° 

100° 

Pb(NO3)2  

36 

44 

51 

56 

79 

127 

Procedure:  Take  56  grams,  or  J  F.W.,  of  litharge,  PbO.  Calcu- 
late the  amount  of  nitric  acid  which  would  be  necessary  to  convert 
it  into  lead  nitrate  and  the  amount  of  water  needed  to  dissolve  the 
salt  thus  formed.  Proceed  to  prepare  lead  nitrate,  striving  to 
obtain  good  crystals  of  as  large  a.  size  as  possible. 

.The  solution  which  is  set  to  crystallize  should  be  slightly  acid,  — 
enough  to  redden  litmus.  If  insufficient  nitric  acid  was  used,  the 
excess  of  PbO  would  have  dissolved  somewhat  in  the  hot  con- 
centrated Pb(NO3)2  solution  forming  the  basic  salt  PbOH.N03 
which  would  separate  as  a  fine  granular  or  flaky  precipitate  when 
the  solution  cooled. 


LEAD   DIOXIDE  249 

QUESTIONS 

1.  Explain  why  lead  nitrate  should  be  less  soluble  in  dilute 
nitric  acid  than  in  pure  water. 

2.  Add  a  few  drops  of  ammonium  hydroxide  to  1  cc.  of  lead 
nitrate  solution.     Then  add  an  excess  of  the  reagent.     Repeat, 
using  sodium  hydroxide  instead  of  ammonium  hydroxide.     Give 
equations  and  explain  the  amphoteric  character  of  lead  hydroxide. 

3.  Are  lead  salts  (nitrate  or  chloride)  appreciably  hydrolyzed 
in  aqueous  solution?     Compare  the  basic  strength  of  the  hydroxide 
of  divalent  lead  with  that  of  aluminum  hydroxide. 

4.  Precipitate  a  little  lead  chloride  by  adding  hydrochloric  acid 
to  a  solution  of  lead  nitrate.     Describe  its  properties  and  compare 
them  with  those  of  lead  tetrachloride  (reference  books).     To  what 
oxide  of  lead  does  lead  tetrachloride  correspond? 

PREPARATION  47 
LEAD  DIOXIDE 

The  compounds  of  lead  in  which  its  valence  is  II  are  the  most 
stable,  but  with  strong  oxidizing  agents  the  valence  may  be  raised 
to  IV.  The  oxide  PbO2  is  very  much  less  basic  than  the  lower 
oxide,  and,  furthermore,  it  is  very  insoluble,  —  either  the  anhy- 
drous oxide  or  its  hydrated  forms,  Pb(OH)4  or  PbO(OH)2.  The 
effect  usually  observed  when  a  salt  of  divalent  lead  is  oxidized,  is 
a  precipitation  of  dark  brown  lead  dioxide.  This  precipitate  can 
be  obtained  by  oxidizing  an  alkaline  solution  containing  a  lead 
salt  with  chlorine,  but  it  cannot  be  obtained  in  an  acid  solution 
with  this  oxidizing  agent  because  hydrochloric  acid  reduces  lead 
dioxide  (see  Exps.  10  and  6,  Chap.  IV).  In  an  acid  solution  con- 
taining no  reducing  agent,  that  is  in  a  nitric  acid  or  a  sulphuric 
acid  solution,  lead  dioxide  can  be  formed  by  the  action  of  a  very 
strong  oxidizing  agent  as  for  example,  by  the  electrolytic  oxidizing 
action,  at  the  anode  of  a  lead  storage  battery. 

In  this  preparation  we  shall  make  use  of  bleaching  powder  in 
a  slightly  alkaline  solution  as  the  oxidizing  agent.  This  is  chosen 
in  preference  to  chlorine  because  it  is  easier  to  handle  and  no 
precautions  need  be  taken  to  avoid  escape  of  objectionable  chlorine 
into  the  laboratory.  It  should  be  recalled  that  the  effect  of  bleach- 
ing powder  is  the  same  as  that  which  would  be  obtained  by  passing 


250  ELEMENTS  OF  GROUP  IV 

chlorine  into  an  alkali.  The  precipitate  finally  obtained  after 
the  bleaching  powder  has  acted  contains  the  greater  part  of  the 
lead  in  the  form  of  lead  dioxide;  but  it  may  also  contain  a  small 
residue  of  unoxidized  lead,  as  Pb(OH)2,  as  well  as  calcium  hydrox- 
ide and  calcium  carbonate  frona  the  bleaching  powder,  and  possibly 
the  salts  calcium  plumbate  and  calcium  plumbite,  CaPbO3  and 
CaPb02.  By  treating  this  precipitate  with  nitric  acid  every- 
thing except  the  lead  dioxide  is  dissolved ,. or  decomposed,  and 
practically  pure  lead  dioxide  remains. 

The  packages  of  bleaching  powder  are  labelled  with  the  percent 
of  available  chlorine.  This  is  the  percent  by  weight  of  chlorine 
which  would  be  given  of  if  the  material  were  treated  with  dilute 
H2S04. 

CaOCl2  +  H2S04  -*  CaS04  +  H2O  +  C12 

It  is  to  be  noted  that  bleaching  powder  may  contain  unavailable 
chlorine  (e.g.,  CaCl2).  Calculate  the  weight  of  bleaching  powder 
with  the  given  content  of  available  chlorine  (assume  30  percent 
if  the  package  is  not  marked)  that  would  be  required  to  oxidize 
the  lead  acetate  used  for  this  preparation. 

Materials:    lead  acetate,  Pb(C2H3O2)2.3H20,  95  grams  =  J  F.W. 

bleaching  powder  containing  30  percent  available 
chlorine,  66  grams. 

6-n  HNO3,  250  cc. 
Apparatus:  8-inch  porcelain  dish. 

2-liter  common  bottle. 

suction  filter. 

Procedure:  Dissolve  the  lead  acetate  in  200  cc.  of  cold  water  in 
the  8-inch  dish;  add  a  solution  of  the  sodium  hydroxide  in  100  cc. 
of  water,  stirring  well,  and  into  the  mixture,  which  should  not  be 
warmer  than  30°,  stir  a  paste  made  by  rubbing  66  grams  of  bleach- 
ing powder  in  a  mortar  with  a  little  water.  Warm  the  mixture 
slowly  to  the  boiling  temperature,  stirring  frequently,  and  finally 
boil  it  for  10  minutes.  Transfer  the  contents  of  the  casserole  to 
a  2-liter  common  bottle  and  wash  the  precipitate  by  decantation 
with  cold  water  (see  Note  5  (b)  on  page  13)  until  the  wash  water 
gives  but  a  slight  test  for  chlorine  ions.  Then  transfer  the  re- 
maining slime  again  to  the  dish,  add  250  cc.  of  6-n  HNO3,  and  boil 
it  for  10  minutes.  Wash  the  residue  of  lead  dioxide  by  decantation 


RED  LEAD  251 

until  the  wash  water  is  no  longer  acid;  transfer  the  product  to  a 
filter  and  let  it  drain  without  suction  (Note  4  (c)  on  page  9). 
After  the  lead  dioxide  has  drained,  remove  the  filter  and  contents 
carefully  from  the  funnel,  unfold  the  filter,  and  spread  it  on  paper 
towels  on  the  steam  table  to  dry.  When  completely  dry,  detach 
the  lumps  of  lead  dioxide  from  the  paper,  and  pulverize  them  in 
a  mortar.  Put  up  the  product  in  a  2-ounce  cork-stoppered  bottle. 

QUESTIONS 

1.  Dissolve  0.1  gram  of  lead  nitrate  in  1  cc.  of  water  and  add 
5  cc.  of  16-n  HNO3.     What  is  the  white  precipitate.     Explain 
the  application  of  the  solubility  product  principle.     Note  partic- 
ularly that  nitric  acid  does  not  oxidize  divalent  lead. 

2.  Show  the  relation  among  ortho-plumbic  acid,  meta-plumbic 
acid,  and  lead  dioxide.     Give  the  symbol  of  sodium  meta-plum- 
bate;   of  calcium  ortho-plumbate. 

3.  Why  could  not  lead  dioxide  be  prepared  equally  well  by 
treating  a  solution  of  lead  chloride  with  chlorine? 

4.  Compare  the  reaction  of  lead  dioxide  and  of  lead  monoxide 
upon  hydrochloric  acid. 

5.  Compare  the  action  of  lead  dioxide  upon  hydrochloric  acid 
with  that  of  manganese  dioxide. 

6.  Why  should  not  lead  dioxide  and  manganese  dioxide  dissolve 
in  dilute  nitric  acid  as  well  as  in  hydrochloric  acid? 

7.  Treat  J  gram  samples  of  lead  dioxide  in  separate  test  tubes 
with  5  cc.  of  16-n  HN03  and  5  cc.  of  36-n  H2SO4.     What  is  the 
gas  evolved  and  what  is  the  white  residue  in  each  case? 

PREPARATION  48 
RED  LEAD,  Pb304 

Of  the  oxides  of  lead,  the  monoxide  PbO  is  the  most  stable  when 
heated  to  a  high  temperature,  and  in  fact  all  of  the  other  oxides 
are  converted  into  this  one  when  they  are  heated  strongly  in  con- 
tact with  the  air.  At  a  moderate  heat,  however,  the  monoxide 
is  capable  of  taking  on  more  oxygen  from  the  air  until  the  com- 
position approximates  that  of  the  formula  Pb3O4.  This  substance 
is  not  to  be  regarded  as  a  simple  oxide  of  lead,  but  rather  as  a 
compound  of  PbO  and  Pb(>2,  in  which  the  monoxide  is  the  basic 


252  ELEMENTS  OF  GROUP  IV 

component  and  the  dioxide  the  acidic.  It  may  thus  be  regarded 
as  the  salt,  lead  orthoplumbate,  2PbO.Pb02  =  Pb2(Pb04).  This 
view  is  strengthened  by  the  behavior  of  the  substance  when  treated 
with  nitric  acid  —  part  of  the  lead  dissolves  to  give  lead  nitrate, 
while  the  other  part  is  left  as  lead  dioxide, 

Pb2(Pb04)  +  4HN03  ->  2Pb(N03)2  +  H4(Pb04) 
H4PbO4  ->  Pb02  +  2H2O. 

The  following  procedure  should  yield  a  product  of  nearly  the 
composition  Pb3O4.  This  substance,  under  the  commercial  name 
of  minium,  finds  use  as  a  red  pigment. 

Procedure:  Spread  25  grams  of  lead  monoxide  in  a  thin  layer 
on  an  iron  or  aluminum  plate  2-4  mm.  thick.  Either  use  the 
variety  of  lead  oxide  which  has  not  been  fused  and  is  known  under 
the  name  of  massicot,  or  use  lead  carbonate,  which  on  being  heated 
yields  a  very  pure  and  finely  divided  lead  monoxide.  Heat  the 
lead  oxide  over  a  ring  burner  so  adjusted  that  the  flames  do  not 
quite  touch  the  metal  plate.  The  latter  must  be  kept  just  below 
a  perceptible  red  heat.  Continue  the  heating  for  6  hours  or  more 
and  turn  over  the  powder  frequently  with  an  iron  spatula.  When 
the  change  is  complete,  the  product  is  dark  brown  when  hot,  a 
bright  scarlet-red  when  partly  cooled,  and  a  somewhat  less  brilliant 
red  when  entirely  cold. 

QUESTIONS 

1.  If  it  is  assumed  that  Pb2O3  and  Pb3O4  are  lead  metaplumbate 
and  lead  orthoplumbate,  respectively,  write  formulas  to  express 
these  facts.    Write  the  formulas  of  the  corresponding  meta-  and 
ortho-plumbic  acids. 

2.  Boil  a  little  of  the  red  lead  with  nitric  acid.     What  is  the 
residue,  and  what  soluble  salt  is  formed?     Test  the  solution  by 
diluting  and  adding  a  few  drops  of  sulphuric  acid. 

•  3.  Heat  a  little  of  the  red  lead  .to  a  dull  red  heat  on  a  thin  piece 
of  iron. 

Experiments 

1.  Carbon  Dioxide.  From  a  generator  (Note  13  (aj,  page 
19)  fill  with  carbon  dioxide  several  test  tubes  inverted  in  a 
pan  of  water,  and  use  them  in  the  following  experiments. 


EXPERIMENTS  253 

(a)  Place  the  thumb  over  the  mouth  of  a  test  tube  of 
carbon  dioxide  and  transfer  it  to  a  beaker  of  freshly  drawn 
water.  Clamp  it  in  position  and  note  the  level  of  the  water 
at  intervals  of  about  15  minutes. 

(6)  Stick  a  gummed  label  on  a  second  test  tube  of  the  gas, 
and  make  a  light  pencil  mark  at  the  middle  point  of  the 
length  of  the  tube.  Place  the  thumb  over  the  mouth  of  this 
tube,  remove  it  from  the  water,  and  turn  it  upright.  Draw 
some  fresh  water  from  the  tap,  and,  removing  the  thumb 
sufficiently,  pour  water  into  the  tube  until  the  level  stands 
at  the  pencil  mark.  Close  the  tube  again  tightly  with  the 
thumb,  invert  the  tube,  and  place  a  heavy  mark  at  the  level 
of  the  surface  of  the  water,  calling  this  mark  1  (it  will  of  course 
nearly  if  not  quite  coincide  with  the  light  mark).  Shake 
the  tube  vigorously  for  sixty  seconds,  then  place  the  mouth 
of  the  tube  under  water  in  the  pan,  remove  the  thumb  and 
mark  the  level  to  which  the  water  rises,  calling  this  mark  2. 
Now  measure  with  a  graduate,  first,  the  volume  between  the 
closed  end  of  the  tube  and  mark  1,  thus  giving  the  volume 
of  carbon  dioxide  taken;  second,  the  volume  between  marks 
1  and  2,  thus  giving  the  volume  of  carbon  dioxide,  measured 
under  atmospheric  pressure,  dissolved  by  the  water;  third, 
the  volume  between  mark  1  and  the  open  end  of  the  tube, 
thus  giving  the  volume  of  water  in  which  the  carbon  dioxide 
was  dissolved.  Take  the  temperature  of  the  water  in  the 
pan.  The  undissolved  gas  was  under  a  pressure  less  than 
atmospheric  just  before  the  thumb  was  removed  from  the 
end  of  the  tube.  Calculate  this  pressure  according  to  Boyle's 
law  from  the  volume.  Then  calculate  the  volume  of  carbon 
dioxide  that  would  be  dissolved  at  the  temperature  of  the 
experiment  in  1  volume  of  water  if  the  gas  were  at  atmos- 
pheric pressure,  applying  Henry's  law  that  the  quantity  of 
a  gas  dissolved  in  a  liquid  is  proportional  to  the  pressure. 

At  15°  1  volume  of  water  will  dissolve  1  volume  of  carbon 
dioxide  under  1  atmosphere  pressure. 

(c)  Take  a  third  tube  of  the  gas,  introduce  5  cc.  of  normal 
NaOH,  and  shake  as  in  (6).  Note  that  except  for  a  small 
bubble,  which  is  doubtless  air  introduced  when  the  NaOH  was 
poured  in,  the  gas  is  entirely  dissolved  by  the  solution. 


254  ELEMENTS  OF  GROUP  IV 

The  carbonic  acid,  which  in  (6)  comes  to  equilibrium  with  the 
carbon  dioxide  in  the  gas  phase,  is  in  this  experiment  neutralized 
by  the  base, 

H2CO3  +  2NaOH  ->  Na^CO.  +  2H2O, 


and  since  5  cc.  of  n-NaOH  reacts,  according  to  the  equation,  with 
carbonic  acid  equivalent  to  56  cc.  of  carbon  dioxide,  all  of  the  gas 
is  dissolved. 

(d)  Repeat  (c)  using  5  cc.  of  normal  Na^COs  instead  of 
NaOH.     Note  that  the  gas  is  nearly  if  not  all  dissolved.     If 
the  experiment  is  repeated  and  the  shaking  is  continued  for 
a  longer  time  the  gas  is  entirely  dissolved. 

According  to  the  equation,  Na^COs  +  H2CO3  —  >  2NaHC03, 
56  cc.  of  carbon  dioxide  should  be  dissolved  in  this  experiment. 

(e)  Repeat    (c),   using    10   cc.    of   lime   water    (saturated 
Ca(OH)2  solution)  instead  of  5  cc.  of  normal  NaOH.     Note 
the  appearance  of  the  solution  during  and  after  the  absorption. 

Write  equation  to  account  for  this  appearance.  What  volume 
of  CO2  should  10  cc.  of  saturated  Ca(OH)2  be  able  to  absorb  (the 
Ca(OH)2  is  0.02  molal)  if  (1)  CaCO3  is  formed,  (2)  Ca(HC03)2  is 
formed? 

(/)  Bubble  carbon  dioxide  slowly  into  a  test  tube  of  lime 
water.  A  white  precipitate  forms  at  once  but  very  soon 
redissolves  and  the  solution  then  remains  clear. 

As  long  as  the  base  is  in  excess  both  stages  of  the  ionization  of 
carbonic  acid  can  proceed  to  completion  because  of  the  removal 
of  H+  ions.  (See  Ionization  of  Polybasic  Acids,  Chap.  Ill,  p.  102), 

Ca++         2OH- 
H2C03  ->  H+  +  HC03-  ->  CO,—       2H+ 

I  I 

CaC03  1     2H20 

but,  as  soon  as  the  carbon  dioxide  is  in  excess  the  concentration 
of  H+-ions  from  the  first  hydrogen  of  the  carbonic  acid  is  greater 
than  can  exist  in  equilibrium  with  the  CO3  —  ions  in  a  saturated 
solution  of  calcium  carbonate;  hence  the  carbonate  ions  are  re- 
moved progressively  and  the  solid  continues  to  dissolve  until  the 
solution  is  again  clear: 


EXPERIMENTS  255 


CaCO3      -* 
H2CO3     ^ 


Ca++ 
HC03- 


co3— 

H+ 

I 
'HCO,~{ 


The  components  boxed  in  by  the  dotted  lines  represent  the  ionized 
salt  calcium  bicarbonate  present  in  the  final  solution. 

2.  Combustibility  of  Carbon  Compounds.  Place  succes- 
sively a  few  drops  of  gasoline,  carbon  disulphide  (€82),  carbon 
tetrachloride  (CCU),  and  chloroform  in  a  porcelain  dish  and 
apply  a  lighted  match. 

Make  mixtures  of  gasoline  and  carbon  tetrachloride  in  the 
proportion  of  5  cc.  and  1  cc.;  4  cc.  and  2  cc.;  3  cc.  and  3  cc.; 
2  cc.  and  4  cc.;  1  cc.  and  5  cc.,  and  under  the  hood  apply  a 
lighted  match  to  each.  If  the  mixture  does  not  at  once  catch 
fire,  heat  it  to  its  boiling  point  and  try  again. 

Place  1  cc.  of  gasoline  and  5  cc.  of  water  in  the  dish  and 
apply  a  lighted  match. 

Gasoline  is  a  mixture  of  hydrocarbons,  that  is  compounds  of 
carbon  and  hydrogen,  of  which  hexane,  C6Hi4,  heptane,  CrHie, 
and  octane,  C8Hi8,  are  the  principal  ones.  Gasoline  and  carbon 
disulphide  are  very  combustible;  the  affinity  of  carbon  for  hydro- 
gen and  for  sulphur  is  not  extremely  great  and  both  hydrogen  and 
sulphur,  as  well  as  carbon  have  a  great  affinity  for  oxygen.  Chlo- 
rine on  the  other  hand  will  not  form  stable  compounds  with  oxygen, 
and  it  does  form  very  stable  compounds  with  carbon.  Carbon 
tetrachloride  and  chloroform  do  not  burn. 

Carbon  tetrachloride  and  gasoline  are  mutually  soluble  in  each 
other  in  all  proportions.  The  vapor  pressure  of  gasoline  is  lowered 
by  the  admixture,  and  furthermore  the  vapor  that  escapes  is  mixed 
with  non-combustible  carbon  tetrachloride  vapor;  it  is  thus 
understandable  that  the  combustibility  of  gasoline  is  lessened  by 
large  admixture  with  this  substance.  Carbon  tetrachloride  is 
extremely  poisonous  and  any  advantage  it  may  have  as  a  fire 
extinguisher  is  neutralized  by  the  great  danger  attending  its 
use. 

Gasoline  and  water  do  not  mix  but  the  former  will  float  in  a 
layer  over  the  latter;  water  is  of  little  effect  in  extinguishing  a 
gasoline  fire. 


256  ELEMENTS  OF  GROUP  IV 

3.  Carbon  Monoxide.     Place  a  plug  of  shredded  asbestos 
loosely  2  inches  from  one  end  of  a  piece  of  difficultly  fusible 
glass  tube.     Fill  the  tube  with  granulated  (not  powdered) 
charcoal  for  a  length  of  about  4  inches  and  insert  another 
plug  of  asbestos:     Join  tl#s  tube  to  a  carbon  dioxide  gener- 
ator, and  let  the  gas  flow  until  air  is  completely  expelled  from 
the  apparatus.     Draw  out  the  end  of  a  delivery  tube  to  a  fine 
capillary  in  order  that  it  may  deliver,  very  small  bubbles 
of  gas,  and  connect  this  by  means  of  a  rubber  tube  with  the 
other  end  of  the  tube  containing  the  charcoal.     Place  a  little 
6-normal  NaOH  in  a  shallow  dish  and  fill  a  test  tube  with  the 
same  solution,  and  invert  the  tube  in  the  solution  in  the  dish. 
Have  the  generator  delivering  a  very  slow  stream  of  carbon 
dioxide,  and  insert  the  delivery  tube  under  the  mouth  of  the 
test  tube.     The  carbon  dioxide  should  be  completely  ab- 
sorbed in  rising  through  the  NaOH  solution.     If  it  is  not,  it 
must  not  be  delivered  so  rapidly  by  the  generator.     Now 
heat  the  charcoal  as  hot  as  possible,  using  the  flame  spreader, 
and  note  if  now  any  gas  issues  from  the  delivery  tube  which 
is  not  absorbed  by  the  NaOH.     Test  the  gas  for  its  com- 
bustibility. 

Carbon  dioxide  is  reduced  by  hot  carbon  to  carbon  monoxide 
which  is  not  an  acidic  oxide  because  it  does  not  react  with  the  base. 
Carbon  monoxide  is  very  combustible  which  fact  may  be  attri- 
buted to  the  tendency  of  carbon  to  return  to  its  usual  valence  of  IV. 

4.  Carbides,     (a)  The  aluminum  nitride  made  in  Prep.  13 
contains   a   considerable   amount   of   carbide   A14C3.     Treat 
some  of  this  product,  or  some  commercial  aluminum  carbide, 
with  6-n  NaOH  in  a  test  tube  with  a  delivery  tube.     Collect 
some  of  the  gas  over  water  which  will  dissolve  all  of  the 
ammonia.     Test  the  combustibility  of  this  gas  and  find  that 
it  burns  with  a  nearly  colorless  flame. 

(6)  Drop  a  little  calcium  carbide,  CaC2,  into  water  in  a  test 
tube,  note  that  a  gas  is  evolved  and  that  this  gas  will  burn 
with  an  intensely  luminous  and  very  smoky  flame. 

Nearly  all  of  the  metals  form  carbides  which  will  hydrolyze  more 
or  less  readily.  The  hydrolysis  product  of  aluminum  carbide  is 
methane 

A14C3  +  12H2O  -»  4A1(OH)3  +  3CH4 


EXPERIMENTS 


257 


from  which  it  is  concluded  that  in  aluminum  carbide  the  carbon 
is  acting  as  a  simple  negative  radical  with  a  valence  of  IV.  Note 
that  the  maximum  negative  valence  of  carbon  is  IV  as  well  as  the 
maximum  positive  valence  (in  CO2)  and  the  arithmetical  sum  of 
the  maximum  positive  and  negative  valences  is  8.  No  element 
is  known  for  which  this  sum  exceeds  8,  and,  with  a  large  number, 
the  value  of  8  is  equalled. 

The  hydrolysis  product  of  calcium  carbide  is  acetylene 
CaC2  +  2H2O  ->  Ca(OH)2  +  C2H2 

The  weight  of  22.4  liters  of  acetylene  is  26  grams  which  is  equal 
to  the  formula  weight  of  C2H2.  It  thus  appears  that  there  is 
present  in  calcium  carbide  the  complex  carbon  radical  C2  with  a 
negative  valence  of  II  for  the  whole  radical.  (Compare  peroxides.) 

5.  Luminosity  of  Flame,  (a)  Note  the  color  of  the  Tirrill 
or  Bunsen  flame  both  when  the  air  inlet  at  the  bottom  is  open 
and  closed.  The  following  experi- 
ments throw  some  light  on  the  cause 
of  the  difference. 

(6)  Fit  up  a  gas  holder  as  shown 
in  the  diagram.  Through  a  allow 
illuminating  gas  to  pass  in  until  the 
bottle  is  one-half  full.  Then  pass  in 
carbon  dioxide  until  the  bottle  is 
filled.  Agitate  the  bottle  enough  to 
ensure  thoroughly  mixing  the  gases. 
Attach  the  capillary  jet  b  to  the 
delivery  tube,  allow  the  gas  to  issue 
from  the  jet,  light  it,  and  observe 
the  color  of  the  flame. 

Repeat  with  varying  proportions 

of  illuminating  gas  and  carbon  di- 

•     oxide,   finding  what   proportions   of 

the  latter  is  sufficient    to    give    a 

colorless  flame. 

(c)  Fill  the  gas  holder  with  acety- 
lene.    Note  the  color  and  character  of  the  acetylene  flame. 

(d)  Repeat  (6),  using  acetylene  in  place  of  illuminating 
gas. 

(e)  The  object  of  this  part  of  the  experiment  is  to  show 


FIG.  21 


258  ELEMENTS  OF  GROUP  IV 

the  effect  of  admixture  with  an  inert  gas  upon  the  decompo- 
sition of  illuminating  gas  or  acetylene. 

Fill  the  gas  holder  with  the  mixture  to  be  tested.  Attach  a 
piece  of  infusible  glass  tube  to  a,  and  at  the  other  end  attach  a 
rubber  connector  with  a  pinejicock  and  the  capillary  6.  Open 
the  pinchcock  until  the  glass  tube  is  full  of  the  gas  mixture. 
Then  heat  the  tube  as  strongly  as  possible,  every  now  and 
then  opening  the  pinchcock  for  a  moment.. 

Test  in  this  way  whether  illuminating  gas  and  acetylene 
are  decomposed  by  heat.  What  is  the  solid  product  of  such 
a  decomposition?  How  does  the  presence  of  some  of  this 
solid  product  affect  the  luminosity  of  a  flame?  If  the  gases 
alone  are  thus  decomposed,  find  if  they  are  or  are  not  so  de- 
composed when  admixed  with  C02  in  the  proportion  that  gave 
a  colorless  flame. 

It  is  believed  that  the  luminosity  of  a  gas  or  candle  flame  is 
caused  by  incandescent  particles  of  carbon  within  the  hot  part 
of  the  flame,  first,  because  when  a  cold  object  is  held  in  a  luminous 
flame  soot  (finely  divided  carbon)  is  deposited,  whereas  the  cold 
object  held  in  the  non-luminous  flame  does  not  become  sooted; 
second,  because  hot  gases  are  known  not  to  emit  light  whereas 
hot  solid  substances  do  give  off  light. 

The  presence  of  a  very  moderate  proportion  of  inert  gas  in 
illuminating  gas  suffices  to  render  its  flame  non-luminous.  Nitro- 
gen exerts  the  same  effect  as  carbon  dioxide. 

Acetylene  burns  with  a  very  much  more  brilliant  flame  than 
illuminating  gas  and  it  takes  a  larger  proportion  of  inert  gas  to 
take  away  its  luminosity. 

Experiment  (e)  shows  that  acetylene  heated  alone  out  of  contact 
with  air  is  very  easily  decomposed  giving  a  dense  deposit  of  carbon 
inside  the  tube;  admixture  of  inert  gas  lessens  the  ease  with  which 
the  carbon  is  deposited.  Illuminating  gas  does  not  give  any 
deposit  of  carbon  within  the  tube  unless  the  tube  is  heated  very 
hot. 

The  facts  above  enumerated  all  strengthen  the  theory  that 
luminosity  is  due  to  the  particles  of  carbon  produced  by  the  de- 
composition of  the  gases  by  the  heat  of  the  flame.  These  particles 
heated  to  incandescence,  give  the  luminosity  until  they  reach  the 
outer  part  of  the  flame  where  the  oxygen  of  the  air  burns  them. 


EXPERIMENTS  259 

When  a  flame  smokes  the  carbon  particles  get  cooled  to  below  the 
kindling  temperature  before  they  come  in  contact  with  sufficient 
oxygen  for  combustion. 

The  question  now  arises  why  mixture  with  an  inert  gas  should 
stop  the  formation  of  the  particles  of  free  carbon.  One  obvious 
reason  is  that  the  inert  gas  absorbs  part  of  the  heat  of  the  flame 
and  thus  prevents  a  temperature  high  enough  to  decompose  the 
hydrocarbon  being  reached.  But  this  does  not  explain  why  the 
mixture  of  acetylene  and  carbon  dioxide  does  not  decompose  in 
the  glass  tube  at  the  temperature  that  decomposed  the  acetylene 
alone.  The  explanation  for  this  fact  is  that  the  rate  of  decom- 
position of  the  hydrocarbon  depends  on  its  concentration  (law  of 
molecular  concentration^;  the  presence  of  inert  gas  lessens  the 
concentration,  and  therefore  the  rate  of  decomposition,  of  the 
hydrocarbon. 

To  return  now  to  the  colorless  Bunsen  flame  the  explanation 
of  its  lack  of  luminosity  which  one  would  at  first  be  inclined  to 
advance  is  that  the  air  admitted  at  the  base  burns  the  carbon 
particles  before  they  have  a  chance  to  glow.  When,  however,  we 
consider  the  fact  that  carbon  dioxide,  or  nitrogen  also  prevents 
the  luminosity,  we  see  that  the  diluting  action  of  the  air  also 
must  be  an  important  factor  in  causing  the  absence  of  luminosity. 

6.  Silicon  Dioxide  and  Silicic  Acid,  (a)  Wet  a  little  silicon 
dioxide  (very  finely  powdered  quartz,  or  better  the  product 
of  Prep.  42),  test  with  litmus,  and  note  that  litmus  is  not 
affected.  Add  Na^COs  solution  and  warm  gently,  noting 
that  there  is  no  effervescence. 

(6)  Collect  a  little  of  a  mixture  of  anhydrous  potassium 
carbonate  and  sodium  carbonate  (the  mixture  melts  more 
easily  than  either  salt  alone)  in  a  loop  on  the  end  of  a  platinum 
wire  and  melt  it  in  the  Bunsen  flame  to  a  clear  bead.  Dip 
the  bead  into  powdered  silicon  dioxide  and  melt  it  again. 
Note  that  the  liquid  bead  effervesces  until  the  silica  has  dis- 
solved. 

(c)  Dilute  5  cc.  of  water  glass,  sodium  silicate  solution, 
with  5  cc.  of  water  in  a  beaker,  and  add  6-n  HC1,  drop  by  drop, 
with  stirring,  noting  that  the  solution  coagulates  to  a  stiff 
apparently  dry  jelly. 

Repeat,  diluting  5  cc.  of  water  glass  with  100  cc.  of  water 


260  ELEMENTS  OF  GROUP  IV 

and  noting  that  coagulation  does  not  take  place  on  acidify- 
ing. 

Silicon  dioxide  is  the  anhydride  of  silicic  acid  but  when  it  is 
entirely  dehydrated  its  action  with  water  is  almost  imperceptible. 
It  does  not  react  with  a  solution  of  the  salt  of  the  weak  carbonic 
acid.  Silicic  acid  is  in  fact  a  far  weaker  acid  than  carbonic. 
With  melted  sodium  carbonate  however,  silicon  dioxide  reacts, 
Si02  +  Na2CO3  — >  Na^SiOs  +  C02 1,  but  this  does  not  necessarily 
show  that  silicon  dioxide  is  more  strongly  acidic  than  carbon  di- 
oxide, the  effect  is  due  rather  to  the  greater  volatility  of  the  carbon 
dioxide.  Water  glass  is  obtained  by  dissolving  the  sodium  silicate 
melt  in  hot  water  under  pressure.  From  it  ordinary  acids  dis- 
place the  weak  silicic  acid  in  colloidal  form.  This  colloidal  silicic 
acid  appears  as  a  jelly  in  concentrated  solutions.  In  dilute  solu- 
tion it  remains  dispersed  so  that  its  formation  is  not  apparent. 
When  the  colloidal  silicic  acid  is  heated  it  is  changed  to  the  anhy- 
dride and  the  latter  will  not  again  take  up  water. 

7.  Hydrolysis  of  Stannous  Salts.     Dissolve  \  gram  crys- 
tallized SnCl2.2H20  in  a  few  drops  of  water.     Note  that  a 
clear  solution  can  be  obtained  if  the  preparation  is  fresh. 
Dilute  the  solution  with  water  and  note  the  white  precipitate. 
Add  a  little  HC1  and  note  that  the  precipitate  redissolves. 

Stannous  salts  in  which  tin  displays  the  lower  valence  of  II 
are  derivatives  of  the  hydroxide  Sn(OH)2.  That  salts,  such  as 
the  chloride,  nitrate,  and  sulphate,  can  exist  in  solution  indicates 
that  stannous  hydroxide  is  basic;  but  that  the  salts  hydrolyze 
very  easily,  with  precipitation  of  basic  salt,  SnCl2  +  H20  ^ 
SnOHCl  |  +  HC1,  indicates  that  base  is  a  weak  one. 

8.  Reducing  Action  of  Stannous  Salts,     (a)  Add  SnCl2 
solution  drop  by  drop,  to  2  cc.  of  HgCl2  solution  diluted  with 
10  cc.  of  water.     Notice  the  white  precipitate  which  turns 
gray  and  then  black  with  more  of  the  reagent. 

(6)  To  2  cc.  FeCl3  solution  diluted  with  10  cc.  of  water 
add  SnCl2  until  the  solution  appears  colorless.  Then  test 
for  ferric  ions  by  adding  KSCN  to  a  part  and  note  that  there 
is  no  red  color.  Test  for  ferrous  ions  by  adding  K3Fe(CN)6, 
and  note  the  deep  blue  precipitate. 


EXPERIMENTS  261 

Tin  has  a  marked  tendency  to  develop  the  valence  IV  character- 
istic of  the  group  and  in  consequence  stannous  compounds  are 
strong  reducing  agents. 

SnCl2  +  2HgCl2  -»  SnCl4  +  2HgCl  I 
SnCl2  +  2HgCl  ->  SnCl4  +  2Hg  | 
SnCl2  +  2FeCl3  -»  SnCl4  +  2FeCl2 

9.  Lead  Salts,     (a)  Dissolve  a  little  pure  crystallized  lead 
nitrate  in  water  and  test  with  litmus  noting  that  it  is  not 
affected.     Dilute  the  solution  and   note  that  there  is  no 
precipitate  of  basic  salt. 

(6)  Moisten  some  litharge  (PbO)  with  water  and  test  with 
litmus,  noting  that  the  latter  is  turned  blue.  Boil  the  litharge 
a  few  minutes  with  10  cc.  of  water,  filter,  and  add  hydrogen 
sulphide  water  to  the  nitrate,  noting  a  little  black  precipitate. 

That  lead  monoxide  is  soluble  enough  to  give  a  precipitate  of 
lead  sulphide  and  that  the  solution  is  alkaline  enough  to  color 
litmus  blue  marks  it  as  exceptionally  basic  for  a  heavy  metal 
oxide.  The  absence  of  hydrolysis  of  the  lead  salts  is  a  further 
evidence  of  the  distinctly  basic  character  of  Pb(OH)2. 

10.  Amphoteric  Character  of  the  Hydroxides  of  Tin  and 
Lead.     Dilute  2  cc.  of  normal  SnCl2  with  10  cc.  of  water. 
From  a  10  cc.  graduate  add  6-n  NaOH  noting  the  amount 
required  to  produce  the  maximum  precipitate  and  again  the 
amount  necessary  to  redissolve  the  precipitate. 

Repeat  using  2  cc.  of  normal  Pb(N03)2  instead  of  SnCl2  and 
note  that  a  very  much  larger  volume  of  the  NaOH  is  necessary 
to  redissolve  the  precipitate. 

Pb(OH)2  is  much  more  basic  than  Sn(OH)2;  it  is  correspond- 
ingly more  weakly  acidic  as  is  shown  by  the  greater  excess  of  base 
required  to  convert  it  to  the  soluble  salt. 

H2SnQ2  +  2NaOH  -4  Na^SnOa  +  2H20 
H2PbO2  +  2NaOH  -*  Na^PbO,  +  2H2O 


Sodium  stannite  and  sodium  plumbite  both  hydrolyze  easily  but 
the  latter  much  more  so,  consequently  the  greater  amount  of  base 
to  overcome  its  tendency  to  hydrolyze. 


262  ELEMENTS  OF  GROUP  IV 

11.  Stannic  Acid.     Heat  J  gram  of  tin  in  a  casserole  with 
a  little  16-n  HN03.     Note  that  red  gases  are  evolved,  that 
the  metal  disintegrates,  and  that  a  white  powder  insoluble 
in  the  nitric  acid,  and  later  insoluble  in  water,  is  formed. 

Concentrated  nitric  acid  oxidizes  tin  to  the  dioxide  which  in  a 
hydrated  form  usually  called  meta-stan'nic  acid  (approximately 
H2Sn03)  is  left  as  the  white  insoluble  residue. 

12.  Thio-Salts  of  Tin.     Perform  Exp.  1  under  Prep.  44. 
Stannous  sulphide  does  not  dissolve  in  Na2S  solution.     Addi- 
tion of  sulphur  causes  it  to  dissolve.     Addition  of  HC1  to  the 
solution  produces  a  yellow  precipitate  and  an  evolution  of 
hydrogen  sulphide. 

Sulphur  and  oxygen  are  interchangeable  in  sulphides  and  oxides. 
Metal  oxides  (basic)  and  non-metal  oxides  (acidic)  combine  to 
form  salts.  Likewise  metal  sulphides  may  combine  with  sulphides 
of  weakly  metallic  or  non-metallic  elements  to  form  salts,  the  so- 
called  thio-salts,  or  sulpho-salts.  Thio-salts  of  a  few  of  the  ele- 
ments, notably  tin,  are  very  well  defined.  Stannous  sulphide  does 
not  form  a  thio-salt;  but  addition  of  sulphur  converts  it  to  stannic 
sulphide  which  does  form  a  soluble  thio-salt  with  the  sulphide  of 
an  alkali  metal. 

Na^S  +  SnS2  -»  Na^SnS, 

In  the  same  way  that  the  higher  oxide  SnO2  is  more  acidic  than 
SnO,  the  higher  sulphide  SnS2  is  more  acidic  and  reacts  more  easily 
with  the  basic  sulphide,  Na^S. 

Addition  of  an  acid  displaces  the  very  weak  thio-stannic  acid 
from  its  salt  solution,  2HC1  +  Na2SnS3  -»  2NaCl  +  H2SnS3;  this 
acid  is  very  unstable  and  decomposes  into  stannic  sulphide,  the 
yellow  precipitate,  and  hydrogen  sulphide,  B^SnSs  — >  H2S  f  + 
SnS2 1. 

13.  Lead  Dioxide,     (a)  ,To  2  cc.  of  Pb(N03)2  solution  add 
5  cc.  of  water  and  6-n  NaOH  until  the  precipitate  first  formed 
redissolves.     Then  add  chlorine  water  and  note  the  dark 
brown  precipitate. 

(6)  Collect  the  precipitate  on  a  filter,  wash  it  with  water, 
break  the  tip  of  the  filter  and  wash  the  precipitate  with  a  jet 
from  the  wash  bottle  into  a  test  tube.  Shake  the  test  tube 


EXPERIMENTS  263 

and  divide  the  suspension  equally  among  four  tubes.  Add 
to  the  respective  tubes  (1)  6-n  HN03  (2)  6-n  H2S04  (3)  6-n 
HC1  (4)  6-n  NaOH  and  note  that  nitric  and  sulphuric  acids 
have  no  effect,  hydrochloric  acid  dissolves  the  brown  precip- 
itate with  evolution  of  chlorine,  and  sodium  hydroxide  has 
no  effect. 

Chlorine  in  alkaline  solution  oxidizes  the  divalent  lead  to  tetra- 
valent,  the  latter  appearing  as  the  very  insoluble  brown  Pb02 

Na2PbO2  +  C12  -» 2NaCl  +  PbO2  J, 

This  highest  oxide  of  lead  has  practically  no  basic  properties 
and  it  does  not  react  with  nitric  and  sulphuric  acids.  Neither  are 
its  acidic  properties  highly  developed  for  it  does  not  react  with 
NaOH.  The  action  with  HC1  has  already  been  discussed;  it  de- 
pends on  the  reducing  action  of  this  acid.  The  next  experiment 
however  throws  a  little  more  light  on  this  subject. 

14.  Lead  Tetrachloride.  Cool  to  0°  5  cc.  of  12-n  HC1  in 
a  test  tube  and  keeping  the  solution  cold  add  about  1  gram 
of  dry  lead  dioxide  a  little  at  a  time  with  shaking.  Note  that 
a  clear  yellow  solution  is  formed.  Add  a  drop  or  two  of  this 
solution  to  500  cc.  of  cold  water  in  a  large  beaker  and  note  an 
opalescent  brown  precipitate  which  rather  slowly  becomes 
visible.  Let  the  rest  of  the  yellow  solution  grow  warm  and 
note  that  chlorine  gas  is  evolved  and  that  a  crystalline  white 
precipitate  separates. 

The  yellow  solution  contains  lead  tetrachloride.  It  is  puzzling 
to  explain  why  lead  dioxide  will  not  react  with  two  of  the  strong 
acids  tried  yet  does  react  with  hydrochloric  acid  to  give  what  is 
apparently  a  salt,  PbCl4.  The  explanation  lies  in  the  character  of 
lead  tetrachloride,  which  is  practically  un-ionized,  and  therefore 
is  hardly  to  be  classed  as  a  salt.  In  the  anhydrous  condition  it  is 
a  liquid  like  carbon  tetrachloride,  etc.  Furthermore  it  combines 
with  excess  HC1  to  form  the  complex  acid  H2PbCl6,  of  which  the 
ammonium  salt  (NH4)2PbCl6  can  be  crystallized.  By  comparison, 
if  nitric  acid  reacted  with  lead  dioxide,  the  tetranitrate,  Pb(NO3)4, 
would  be  the  product;  this  presumably  would  be  highly  ionized 
like  all  nitrates,  which  means  that  it  would  have  to  hydrolyze 
completely. 


264  ELEMENTS  OF  GROUP  IV 

With  a  large  amount  of  water  the  lead  tetrachloride  hydrolyzes 

PbCl4  +  4H20  ->  Pb(OH)4 1  +  4HC1 
Pb(OH)4  ->  Pb02  +  2H20 

giving  the  brown  precipitate.  ,^In  concentrated  solution  it  decom- 
poses into  chlorine  and  PbCl2,  the  white  crystalline  precipitate. 

15.  Stability  of  Lead  Carbonate,  (a)  To  a  neutral  lead 
nitrate  solution  add  Na^COa  solution  drop  by  drop,  noting  the 
white  precipitate  and  the  absence  of  effervescence. 

(6)  Heat  a  little  dry  white  lead  carbonate  in  a  test  tube 
and  note  that  it  changes  to  a  yellow  powder  when  it  has 
become  moderately  hot.  The  temperature  is  much  higher 
than  that  required  to  decompose  copper  carbonate.  Treat 
some  of  the  residue  with  dilute  HN03  and  note  that  it  dis- 
solves without  effervescence. 

The  facts  that  lead  carbonate  PbCOs  will  precipitate  without 
hydrolyzing  to  a  basic  carbonate,  and  that  lead  carbonate  must 
be  heated  moderately  hot  to  be  decomposed,  both  confirm  the 
conclusion  already  made  that  PbO  is  a  distinctly  basic  oxide. 

GENERAL  QUESTIONS  IX 

1.  Arrange  a  table  of  the  dioxides  of  the  elements  of  Group  IV: 
column  1,  the  formulas  of  the  dioxides,  placing  those  of  the  A 
family  at  the  left  and  those  of  the  B  family  at  the  right  of  the 
column;   column  2,  the  character  of  the  dioxide  specifying,  a  = 
distinctly  acidic,  w.  a.=  weakly  acidic;  ind.=  indifferent;  w.  b.= 
weakly  basic;    6.=  distinctly  basic;    w.  a.,  w.  b.=  amphoteric; 
column  3,  the  formula  of  as  well  defined  a  salt  as  possible  of  the 
dioxide;  column  4,  the  extent  of  hydrolysis  of  this  salt,  specifying 
much,  little,  or  none. 

2.  Make  a  similar  table  embracing  the  lower  oxides,  CO,  SnO, 
Ce203,  PbO,  Pb304. 

3.  Make  a  table  for  the  tetrachlorides  of  all  of  the  elements  of 
Group  IV:  column  1,  the  formula  of  the  tetrachloride;  column  2, 
its  state  of  aggregation,  specifying,  gas,  liquid,  or  solid;  column  3, 
its  boiling  point  at  atmospheric  pressure,  specifying  dec  if  it  de- 
composes before  the  boiling  point  is  reached;  column  4,  the  equa- 
tion for  its  reaction  with  a  large  amount  of  water. 

Judging  from  the  decreasing  metallic  properties  in  the  series, 


GENERAL  QUESTIONS  IX  265 

Pb,  Sn,  Si,  C,  we  should  expect  the  tetrachlorides  to  hydrolyze 
more  readily  as  we  progress  in  this  order.  Two  factors  modify 
this  effect,  the  tendency  to  form  a  complex  acid  such  as  H2SnCl6 
with  the  anion  SnCl6  ,  and  the  insolubility  or  total  lack  of  ioniza- 
tion  of  the  tetrachloride.  Explain  from  this  point  of  view  why 
carbon  tetrachloride  and  carbon  disulphide  are  without  perceptible 
action  with  water. 

4.  Find  out  what  elements  of  Group  IV  form  carbonates  and 
give  the  formulas  of  the  carbonates  and  approximately  their 
relative  stability. 

5.  What  is  a  thio-salt?     Describe  how  a  thio-salt  of  tin  can  be 
formed,  and  discuss  its  properties  and  its  relation  to  the  corre- 
sponding oxy-salt. 


CHAPTER  X 

ELEMENTS  OF  THE  GROUP  V  OF  THE 
PERIODIC   SYSTEM 

In  this  group,  as  in  Groups  III  and  IV,  the  difference  in  prop- 
erties between  the  elements  of  Families  A  and  B  is  not  so  striking 
as  in  Groups  I  and  II  (or  as  in  Groups  VI  and  VII),  and  the  whole 
group  is  considered  under  the  same  heading.  But  it  is  also  true 
that  the  elements  of  Family  A,  that  is,  vanadium,  columbium,  and 
tantalum,  are  of  comparatively  infrequent  occurrence,  and  are 
given  no  attention  in  this  course.  On  the  other  hand  all  of  the 
elements  of  Family  B  are  of  frequent  occurrence  and  considerable 
importance. 

The  characteristic  valence  of  the  group  is  five,  corresponding  to 
the  oxide  M2O5,  but  the  elements  likewise  exhibit  a  valence  of  three 
in  the  oxide  M20s.  It  is  noteworthy  that  the  valence  is  nearly 
always  either  three  or  five. 

It  is  tr.ue  in  this  group,  as  well  as  in  Group  IV,  that  the  acid- 
forming  properties  are  most  marked  in  the  elements  of  low  atomic 
weight  (nitric  acid  is  one  of  the  strongest  acids),  and  decrease  with 
increasing  atomic  weight ;  whereas  the  base-forming  properties  are 
most  strongly  developed  with  the  elements  of  high  atomic  weight. 

PREPARATION  49 
ORTHO-PHOSPHORIC  ACID,  H3PO4 

A  rather  impure  grade  of  phosphoric  acid  can  be  obtained  from 
natural  calcium  phosphate  by  decomposition  with  sulphuric  acid, 
but  a  pure  product  may  be  most  readily  obtained  by  oxidizing 
phosphorous  by  means  of  nitric  acid. 

The  solution  first  obtained  by  the  action  of  dilute  nitric  acid 
upon  phosphorus  contains  a  considerable  quantity  of  phosphorous 
acid,  H3PO3;  but  upon  boiling  down  this  solution,  a  point  is  reached 
at  which  a  rather  strong  reaction  takes  place,  which  consists  of  an 
oxidation  of  the  phosphorous  to  phosphoric  acid  by  means  of  the 
nitric  acid  still  present. 

266 


ORTHO-PHOSPHORIC  ACID  267 

Commercial  phosphorus  often  contains  a  small  quantity  of 
arsenic.  This  on  the  treatment  with  nitric  acid  is  oxidized  to 
arsenic  acid,  which,  unless  removed  by  hydrogen  sulphide,  would 
contaminate  the  preparation  of  phosphoric  acid. 

Phosphorous  acid  may  always  be  present  in  the  product  in  case 
the  oxidation  with  nitric  acid  has  not  been  complete,  and  its 
presence  may  be  detected  by  its  ability  to  reduce  silver  nitrate 
and  give  a  black  precipitate  of  metallic  silver, 

H3PO3  +  2AgN03  +  H2O  =  2HN03  +  2Ag  j  +  H3P04. 

Materials:     red  phosphorus,  P,  31  grams  =  1  F.W. 

6-n  HNO3,  295  cc. 

16-n  HNO3,  20  cc. 

seed  crystal  of  H3PC>4. 
Apparatus:  2-liter  round  bottom  flask. 

condenser  to  hang  in  neck  of  flask  (Fig.  22). 

8-inch  evaporating  dish. 

thermometer. 

100  cc.  casserole. 

2-ounce  glass-stoppered  sample  bottle. 

Procedure:  Place  the  nitric  acid  in  a  2-liter  round-bottom  flask, 
connect  the  condenser  with  the  cold  water  tap  and  hang  it  in  the 
neck  of  the  flask  to  condense  acid  vapors  ^ 

and   allow  them  to  drip  back  into  the  jT~ 

reacting  mixture.  Add  about  one  fifth 
of  the  phosphorus  and  warm  the  flask 
very  cautiously  until  red  vapors  begin  to 
appear;  then  stand  the  flask  in  a  wooden 
ring  and  allow  the  reaction  to  proceed, 
keeping  a  pan  of  cold  water  at  hand, 
in  which  to  immerse  the  flask  if  the  reac- 
tion gets  too  violent.  After  the  foaming 

has  nearly  stopped  add  a  little  more  of  FlG'   22'  ,  Condenf rs   to 
,,          ,         ,  .  .,  ,.,    Hang  in  Mouth  of   Flask 

the    phosphorus,  and    again   wait    until 

*  The  one  at  the  left  is  more  effi- 

actlOn    has    nearly    Stopped,    and    SO    On,    dent;  the  one  at  the  right  may 

until  all  of  the  phosphorus  is  used.     Then  be  *****  in  a  ^minutes  by 

bending  common  delivery  tube. 

add  20  cc.  of  16-n  HN03;   transfer  the 

solution  to  an  8-inch  evaporating  dish  and  boil  it  under  the 

hood  until  a  rather  violent  reaction  begins  to  take  place.     Remove 


268  ELEMENTS  OF  GROUP  V 

the  flame  and  let  the  reaction  proceed;  it  grows  very  violent  and 
then,  in  the  most  remarkable  fashion  ceases  altogether. 

The  solution  should  now  be  clear  and  almost  colorless,  and  should 
contain  no  phosphorous  acid.  If  it  contains  a  fine  black  sus- 
pension, dilute  it  with  an  equal  volume  of  water  and  filter  it  without 
suction.  Test  for  phosphorous 'acid  by  adding  a  few  drops  of  the 
liquid  to  10  cc.  of  water,  adding  1  cc.  of  .05-n  AgNO3  and  warming; 
a  dark  coloration,  or  black  precipitate,  appearing  within  two  min- 
utes indicates  phosphorous  acid.  If  it  is  found,  follow  the  special 
procedure  given  in  Note  1;  otherwise  transfer  the  solution  to  a 
small  casserole  and  evaporate  it  over  a  small  flame  until  a  ther- 
mometer whose  bulb  is  immersed  in  it  stands  at  180°.  During 
this  final  evaporation  one  must  give  it  constant  attention,  for  if 
it  is  left  and  the  temperature  rises  above  180°,  not  only  does  the 
ortho-phosphoric  acid  become  changed  partially  into  pyrophos- 
phoric  acid,  but  it  attacks  very  strongly  the  material  of  the  dish, 
and  the  preparation  becomes  contaminated.  Pour  the  liquid 
while  still  warm  into  a  previously  weighed  2-ounce  glass-stoppered 
bottle,  and  stopper  it  tightly.  When  cool  introduce  a  small 
crystal  of  phosphoric  acid  to  induce  crystallization  of  the  mass. 

Note  1.  If  phosphorous  acid  was  found  at  the  point  where  the 
test  was  made,  it  showed  that  the  reaction  had  been  allowed  to  take 
place  too  violently  in  the  earlier  part  of  the  procedure,  the  heat 
driving  nitric  acid  vapor  out  of  the  flask.  Unless  nitric  acid  is 
present  in  the  right  amount  and  at  the  right  concentration  when 
the  secondary  reaction  takes  place,  the  phosphorous  acid  is  not  all 
oxidized  and  it  is  afterwards  extremely  difficult  to  bring  about 
conditions  under  which  the  oxidation  can  be  completed.  However 
the  following  may  be  tried :  Pour  the  solution  back  into  the  large 
flask,  add  275  cc.  of  6-n  HNO3  and  10  cc.  of  12-n  HC1  to  act  as  a 
catalyzer,  insert  the  condenser  in  the  neck  of  the  flask  and  boil 
very  gently  for  an  hour.  Then  proceed  as  before. 

Note  2.  In  the  above  procedure  no  provision  is  made  for  re- 
moving traces  of  arsenic.  If  this  is  to  be  done  the  solution,  im- 
mediately after  the  violent  secondary  reaction  has  ceased,  is 
poured  into  a  flask,  diluted  to  about  a  liter  with  water,  saturated 
with  hydrogen  sulphide  gas,  stoppered,  and  allowed  to  stand  over 
night.  If,  the  next  morning,  the  contents  of  the  flask  smell 
strongly  of  hydrogen  sulphide,  the  precipitate  of  arsenic  sulphide 
is  filtered  off;  if  not,  the  solution  is  again  treated  with  hydrogen 


CRYSTALLIZED  ARSENIC  ACID  269 

sulphide  in  the  same  manner  as  before.     Evaporate  the  filtrate 
until  its  temperature  has  risen  to  125°,  and  proceed  as  above. 

Note  3.  If  no  crystallized  phosphoric  acid  is  obtainable  the 
sirupy  acid  can  be  made  to  crystallize  spontaneously  if  it  is  placed 
in  a  vacuum  desiccator  over  concentrated  sulphuric  acid  and  cooled 
with  a  freezing  mixture. 

QUESTIONS 

1.  Write  the  reaction  by  which  phosphoric  acid  can  be  prepared 
from  calcium  phosphate. 

2.  How  can  phosphoric  anhydride,  pyrophosphoric  acid,  and 
metaphosphoric  acid  be  prepared?     Give  formulas.     Why  cannot 
the  anhydride  be  prepared  by  heating  ortho-phosphoric  acid? 
For  what  practical  purpose  is  phosphoric  anhydride  used? 

3.  Compare  the  acid  strength  of  phosphoric  acid  with  that  of 
other  common  acids.     Do  all  three  hydrogen  ions  dissociate  with 
equal  readiness? 

4.  Give   the   formulas    of   primary,    secondary,    and   tertiary 
sodium  phosphates.     State  how  the  solution  of  each  behaves  with 
litmus. 

5.  Write  the  reaction  for  the  precipitation  which  occurs  when 
magnesium  chloride  and  a  large  excess  of  NH4OH  are  added  to  a 
solution  of  phosphoric  acid.     This  precipitate  constitutes  one  of 
the  most  important  tests  for  phosphoric  acid. 

6.  Give  an  example  of  phosphorous  acid  acting  as  a  reducing 
agent. 

PREPARATION  50 
CRYSTALLIZED  ARSENIC  ACID  (H3AsO4)2.H2O 

Arsenic  acid  in  its  properties  shows  a  striking  similarity  to 
phosphoric  acid;  and  even  the  method  of  its  preparation  is  similar, 
in  that  use  is  made  of  the  oxidizing  action  of  nitric  acid.  Instead 
of  starting  with  uncombined  arsenic,  however,  use  is  made  of 
arsenious  oxide,  As2O3,  a  product  which  condenses  in  the  flues 
wherever  ores  which  contain  arsenic  are  roasted.  By  the  nitric 
acid  this  is  oxidized  to  the  higher  oxide,  As2O5,  which,  with  water, 
yields  arsenic  acid,  H3AsO4.  By  evaporating  its  solution  for  a 
long  time  on  the  water  bath,  crystals  of  ortho-arsenic  acid  having 
the  composition  HsAsC^,  can  be  obtained.  By  prolonged  evap- 


270  ELEMENTS  OF  GROUP  V 

oration  at  higher  temperatures  crystals  of  the  composition 
and  HAs03,  respectively,  can  be  obtained.  When  a  solution  of 
arsenic  acid  is  boiled  down  according  to  the  following  directions, 
a  liquid  is  obtained  of  almost  exactly  the  composition  given  by  the 
formula  (HsAsO^.H^O. .  This,*  liquid  when  cooled  to  35.5°  or 
below  can  be  crystallized  to  a  solid  product  of  the  same  com- 
position, and  this  is  the  most  satisfactory  form  in  which  to  crystal- 
lize arsenic  acid.  It  is  interesting  to  note  that  this  liquid  can  be 
much  supercooled  below  35.5°,  but  that  when  once  crystallization 
is  induced  the  temperature  immediately  rises  to  this  point  and 
remains  there  until  solidification  is  complete.  Likewise  when  the 
solid  is  being  melted  the  temperature  will  not  rise  above  the  melt- 
ing point,  35.5°,  until  the  whole  mass  is  liquefied. 

Materials:     arsenious  oxide,  As2O3,  50  grams  =  J  F.W. 

16-n  HNO3,  75  cc. 

seed  crystal  of  (HsAsO^.H-jO. 
Apparatus:  500  cc.  casserole. 

150  cc.  casserole. 

thermometer. 

2-ounce  glass-stoppered  sample  bottle. 

Procedure:  Place  50  grams  of  arsenious  oxide  in  a  good-sized 
casserole;  add  20  cc.  of  water,  and  then  at  the  hood  add  75  cc.  of 
16-n  HNOs,  warm  occasionally  just  enough  to  keep  up  an  action, 
but  do  not  allow  the  reaction  to  become  violent,  because  the  heat 
would  drive  off  nitric  acid.  When  red  vapors  cease  to  be  given  off, 
the  original  white  powder  should  have  all  dissolved,  and  a  clear 
colorless  or  very  pale  yellow  solution  should  be  obtained.  It  will 
sometimes  happen  however  for  no  very  apparent  reason  that  the 
reaction  stops  with  a  considerable  amount  of  white  powder  still 
undissolved  even  although  a  plentiful  excess  of  nitric  acid  may  be 
present.  When  this  does  happen  the  addition  of  5  cc.  of  6-n  HC1 
will  make  the  reaction  start  up  vigorously  again  and  run  to  com- 
pletion. Arsenious  chloride  is  volatile  and  very  poisonous;  if 
HC1  is  added  keep  the  dish  under  the  hood  during  the  reaction  and 
the  subsequent  evaporation.  Since  HC1  is  not  an  oxidizing  agent 
its  action  must  be  essentially  that  of  a  catalyzer.  Finally  evap- 
orate the  solution,  holding  the  casserole  with  the  hand  and  rotating 
it  to  spread  the  liquid  up  on  the  sides,  until  the  residue  is  just  dry. 
This  residue  should  be  arsenic  pentoxide  and  it  should  dissolve 


CRYSTALLIZED  ARSENIC  ACID  271 

completely,  although  somewhat  slowly,  when  treated  with  60  cc. 
of  water  (see  Note  1).  Evaporate  the  solution  by  boiling  it  gently 
in  a  small  casserole  until  the  temperature  has  risen  to  115°.  Then 
transfer  the  liquid  to  a  very  narrow  beaker  or  a  test  tube,  and  boil 
it  carefully  with  a  small  flame  until  the  temperature  shown  by  a 
thermometer  inserted  in  the  liquid  has  just  risen  to  160°.  Cool  the 
product  to  below  35.5°,  place  it  in  a  weighed  sample  bottle,  and 
seed  it  with  a  small  crystal  of  (H3AsO4)2.H20,  whereupon  the  whole 
will  slowly  crystallize  to  a  solid  mass.  Stopper  the  bottle  tightly, 
since  arsenic  acid  takes  moisture  rapidly  from  the  atmosphere. 

Note  1.  If  the  residue  on  evaporation  does  not  redissolve  after 
warming  it  ten  minutes  with  60  cc.  of  water,  it  contains  arsenious 
oxide  either  from  incomplete  oxidation  by  nitric  acid,  or  from  a 
decomposition  of  arsenic  pentoxide  by  over  heating  Test  1  cc. 
of  the  suspension  containing  the  undissolved  substance  by  adding 
10  cc.  of  water,  then  solid  sodium  bicarbonate  until  no  more 
effervescence  occurs,  and  then  a  considerable  quantity  in  excess. 
Add  to  this  a  solution  of  iodine,  drop  by  drop.  The  amount  of 
the  latter  which  is  decolorized  (if  any)  corresponds  to  the  amount 
of  arsenious  acid  (As203),  which  was  in  the  sample. 

If  arsenious  acid  is  present  it  must  be  oxidized  by  further  treat- 
ment with  16-n  HN03  and  a  little  HC1 


QUESTIONS 

1.  Compare  the  strength  of  arsenic  and  arsenious  acids.     Of 
what  general  rule  is  this  comparison  an  example? 

2.  To  a  solution  of  arsenic  acid  (0.1  gram  in  10  cc.  of  water) 
add  magnesium  chloride  and  then  NH4OH  until  strongly  alkaline. 
Compare  with  Question  5  under  Phosphoric  Acid. 

3.  Add  a  little  potassium  iodide  solution  to  some  arsenic  acid 
solution,  and  warm  gently.     Is  iodine  set  free?     Write  equation? 

Prepare  a  faintly  alkaline  solution  of  arsenious  acid  as  follows: 
Dissolve  a  minute  quantity  of  arsenious  oxide  in  not  more  than  2 
or  3  drops  of  hydrochloric  acid;  dilute  to  10  cc.  and  add,  without 
heating,  a  considerable  amount  of  sodium  bicarbonate  in  excess 
of  what  is  necessary  to  neutralize  the  acid.  To  this  solution  add, 
drop  by  drop,  a  solution  of  iodine,  and  determine  if  any  free  iodine 
disappears.  Write  the  equation.  So  far  as  the  state  of  oxidation 
of  the  arsenic  is  concerned,  the  reaction  is  exactly  the  reverse  of  the 


272  ELEMENTS  OF  GROUP  V 

one  preceding.  Recall  a  previous  instance  in  which  the  direction 
of  a  reaction  of  oxidation  and  reduction  is  changed  on  passing  from 
an  acid  to  an  alkaline  solution. 

PREPARATION  51 

ANTIMONY  TRICHLORIDE  FROM  STIBNITE 
(BY-PRODUCT:    ANTIMONY  OXYCHLORIDE) 

Native  antimony  sulphide  (stibnite)  dissolves  quite  readily  in 
hydrochloric  acid,  yielding  antimony  trichloride, 

Sb2S3  +  6HC1  =  2SbCl3  +  3H2S. 

If  the  solution  so  obtained  is  distilled,  there  pass  off  at  first 
only  steam  and  hydrochloric  acid,  later  a  mixture  of  hydrochloric 
acid  and  antimony  trichloride,  and  finally  pure  antimony  tri- 
chloride. 

.  Antimony  trichloride  hydrolyzes  with  a  moderate  amount  of 
water,  giving  a  precipitate  according  to  the  reactions, 

SbCl3  +  2H20  =  SbCl(OH)2 1  +  2HC1, 
SbCl(OH)2  =  SbOCl  +  H20; 

with  more  water  a  further  hydrolysis  takes  place: 

4SbOCl  +  H20  =  Sb4O5Cl2  +  2HC1. 

The  product  obtained  in  this  preparation  by  mixing  the  next  to  the 
last  distillates  with  a  considerable  amount  of  water  has  the  latter 
composition.  This  compound,  however,  if  repeatedly  boiled  with 
fresh  portions  of  water  may  be  made  to  undergo  complete 
ysis,  leaving  finally  only  Sb203. 

Pure  antimony  trichloride  melts  at  73°  and  boils  at  223°. 

Materials:     stibnite,  Sb2S3,  168  grams  =  i  F.W. 

12-n  commercial  concentrated  HC1,  840  cc. 

shredded  asbestos  for  filter. 
Apparatus:  8-inch  porcelain  dish. 

suction  filter. 

350  cc.  retort. 

1-liter  flask. 

2-liter  common  bottle. 


ANTIMONY  TRICHLORIDE  FROM  STIBNITE  273 

Procedure:  Treat  the  powdered  stibnite  in  an  8-inch  dish  at  the 
hood  with  the  commercial  hydrochloric  acid;  warm  the  mixture 
slightly  and  keep  it  at  50-70°,  with  frequent  stirring,  for  20  minutes. 
Finally,  boil  the  solution  for  5  minutes.  Then  add  15  cc.  more  of 
concentrated  hydrochloric  acid;  filter  the  solution  through  asbestos 
felt  (Note  4  (d))  which  has  previously  been  moistened  with  hydro- 
chloric acid,  and  rinse  the  residue  onto  the  filter  with  an  additional 
15  cc.  of  hydrochloric  acid.  Evaporate  the  filtrate  in  an  open  dish 
to  200  cc.;  then  transfer  it  to  a  retort,  in  the  bottom  of  which  is 
placed,  to  prevent  bumping,  about  a  teaspoonful  of  small  bits 
cracked  from  an  unglazed  porcelain  dish.  Place  the  retort  on  a 
sand  bath  and  distill,  after  first  covering  the  bulb  of  the  retort  with 
an  asbestos  mantle  to  prevent  loss  of  heat.  At  first  insert  the  neck 
of  the  retort  into  a  liter  flask  half  filled  with  cold  water  (to  absorb 
the  hydrochloric  acid).  When  a  little  of  the  distillate  begins  to  give 
a  precipitate  on  dropping  into  a  tube  of  cold  water,  exchange  the 
receiving  flask  for  a  smaller  dry  one  and  continue  the  distillation 
until  a  drop  of  the  distillate  will  solidify  when  cooled  on  a  watch 
glass.  Save  the  portion  thus  obtained  for  later  use  and  continue 
distilling,  using  a  wide  6-inch  test  tube,  which  has  previously  been 
weighed,  as  a  receiving  vessel,  until  the  liquid  is  all  driven  out  of 
the  retort.  Stopper  the  test  tube  tightly  and  preserve  the  prep- 
aration in  it.  If  the  product  thus  obtained  is  not  white  it  should 
be  c1'  ^olved  in  concentrated  hydrochloric  acid  and  redistilled. 

Note.  In  case  the  stibnite  contains  a  considerable  quantity  of 
silicates  soluble  in  acids,  there  will  be  left  in  the  retort  as  the  dis- 
tillation progresses  a  quantity  of  gelatinous  silicic  acid  which  is 
liable  to  interfere  with  obtaining  distinct  fractions  of  the  distillate. 
In  suck  a  case  distill  until  the  residue  in  the  retort  is  left  dry,  but 
wi  making  the  final  change  in  receiving  vessels.  Then  pour 

all  of  the  distillate  containing  any  of  the  antimony  salt  into  a  fresh 
retort  and  distill  again,  this  time  separating  the  fractions. 

4  riimony  Oxychloride.  —  Pour  the  portion  of  the  distillate 
sa\  rom  the  above  procedure  into  2  liters  of  water.  Stir,  allow 
to  st  'e,  and  draw  off  the  clear  liquid.  Stir  up  with  water  once 
more,  let  settle,  draw  off  as  much  of  the  water  as  possible,  and  drain 
the  precipitate  on  a  suction  filter.  Dry  it  on  paper  towels  and  put 
it  up  in  a  cork-stoppered  test  tube. 


274  ELEMENTS  OF  GROUP  V 

QUESTIONS 

1.  Treat  a  fragment  of  antimony  trichloride  with  water.     Why 
does  it  not  give  a  clear  solution?     Add  HC1.     Why  does  this  cause 
a  clear  solution  to  be  formed?    ± 

2.  Pass  hydrogen  sulphide  itito  the  solution  of  antimony  tri- 
chloride.    What  is  the  color  of  the  precipitate?     How  could  it  be 
converted  into  a  product  like  stibnite?  ^ 

3.  Compare  the  reactions  of  phosphorus,  arsenic,  and  antimony 
trichlorides  with  water.     Is  hydrolysis  more  or  less  complete  in  the 
case  in  which  a  precipitate  forms? 

PREPARATION  52 
SODIUM  SULPHANTIMONATE  Na3SbS4.9H2O 

The  oxides  of  arsenic  and  antimony,  and  more  particularly  the 
higher  oxides,  are  acidic  in  nature,  and  form  salts  with  basic  oxides. 


3Na2O  +  As203  = 

3Na2O  +  As205  =  2Na3As04. 

Sulphur,  in  accord  with  its  similarity  to  oxygen,  can  be  substituted 
for  the  latter  in  many  of  its  compounds  without  essentially  altering 
their  chemical  nature,  and  the  compounds  thus  obtained  have  the 
same  nomenclature  as  the  corresponding  oxygen  compounds,  except 
that  the  syllable  thio  or  sulpho  is  inserted.  Thus  sulpho-salts  are 
produced  in  the  same  manner  as  the  oxy-salts  above: 

As2S3  =  2Na3AsS3; 
As2S5  =  2Na3AsS4. 

The  sulpho-salts  of  arsenic,  antimony,  and  stannic  tin  are  particu- 
larly characteristic  of  these  metals.  (See  Prep.  44,  and  Exp.  12, 
Chap.  IX.)  They  are  easily  produced,  and  all  are  soluble.  They 
are  stable  in  neutral  or  basic  solutions,  but  are  decomposed  by  acids, 
because  the  anions  of  the  salts  combine  with  hydrogen  ions  to 
produce  the  very  weak  sulpho-acids,  which,  being  unstable,  decom- 
pose at  once  into  the  sulphides  of  the  metals  and  hydrogen  sul- 
phide: 

6H+  +  2AsS3-  -  -*  2H3AsS3  ->  3H2S  +  As2S3  J,  ; 

6H+  +  2AsS4  ---  -*  2H3AsS4  -»  3H2S  +  As2S5  \. 


SODIUM  SULPHANTIMONATE  275 

Sodium  sulphantimonate  can  be  prepared  from  stibnite  by  the 
combined  action  of  a  solution  of  sodium  sulphide  and  sulphur, 

2S  +  Sb2S3  ->  Sb2S5, 


it  crystallizes  well  with  nine  molecules  of  water. 

The  hydrolysis  of  this  salt,  which  produces  a  dirty  appearing 
reddish  brown  precipitate  consisting  of  an  indefinite  mixture  of 
Sb2S5  and  Sb205  may  be  prevented  by  an  excess  of  sodium  sulphide 
or  by  the  presence  of  sodium  hydroxide. 

Materials:    powdered  stibnite,  Sb2S3,  67  grams  =  £  F.W. 

sodium  sulphide,   Na2S.9H2O,   140  grams,  or  use 
47  grams  of  anhydrous  sodium  sulphide  and  an 
additional  93  cc.  of  water. 
powdered  sulphur,  13  grams. 
Apparatus:  500  cc.  casserole. 
suction  filter. 

6-inch  crystallizing  dish  with  watch  glass  or  glass 
plate  to  cover  it. 

Procedure:  To  the  powdered  stibnite,  sodium  sulphide,  and 
powdered  sulphur  in  a  casserole  add  150  cc.  of  water,  bring  to  a 
boil,  and  keep  at  the  boiling  temperature  for  15  minutes.  Filter 
with  suction  and  rinse  the  residue  in  the  dish  and  on  the  filter  with 
hot  water,  bringing  up  the  volume  of  the  solution  to  250  cc.  While 
still  hot  put  it  away  in  a  covered  dish,  with  a  towel  placed  over  it, 
to  crystallize.  Drain  the  crystals;  evaporate  the  mother  liquor 
somewhat  to  obtain  a  second  crop  of  crystals.  If  there  is  any 
tendency  for  a  muddy  brownish  precipitate  to  form  in  the  solution, 
or  for  the  same  substance  to  form  as  a  scum  on  the  crystals,  add  a 
little  6-n  NaOH  to  the  solution  and  rinse  the  crystals  in  it.  Spread 
the  crystals  on  paper  towels,  and  stopper  them  tightly  in  an 
8-ounce  cork-stoppered  bottle  as  soon  as  they  are  dry. 

QUESTIONS 

1.  What  is  the  acid  of  which  Na3SbS4  is  the  salt?    Add  HC1  to 
a  solution  of  this  salt.     Is  the  acid  set  free?     Is  it  a  stable  acid? 

2.  What  is  the  primary  reaction  in  the  hydrolysis  of  Na3SbS4? 
What  secondary  reaction  accounts  for  the  formation  of  the  precip- 


276  ELEMENTS  OF  GROUP  V 

itate  if  its  formula  is  Sb2S5?  Write  equation  for  change  of  Sb2Ss 
to  nSb2S5.mSb205.  Explain  why  both  Na2S  and  NaOH  are  capable 
of  repressing  the  hydrolysis. 

3.  What  general  principle  is  illustrated  in  the  fact  that  Sb2Ss 
dissolves  much  less  readily  than^Sb2S5  in  Na^S  solution? 

PREPARATION  53 
ANTIMONY  PENTASULPHIDE,  Sb2S5 

This  compound  cannot  be  prepared  directly  from  the  trisulphide 
and  sulphur,  because  it  is  decomposed  at  a  temperature  below  that 
at  which  the  latter  substances  would  react.  As  has  just  been  seen, 
however,  the  higher  sulpho-salt  of  antimony  can  be  readily  pre- 
pared in  the  wet  way;  and  this,  on  decomposition  with  a  dilute 
acid,  yields  antimony  pentasulphide.  This  substance  is  much 
used  in  vulcanizing  rubber. 

Materials:    sodium  sulphantimonate,  Na3SbS4.9H2O  from  pre- 
ceding preparation,  48  grams  =  TV  F.W. 

6-n  H2S04,  108  cc. 
Apparatus:  large  common  bottle,  2  liters  or  larger. 

5-inch  funnel  and  plain  10-inch  filter. 

Procedure:  Dissolve  the  sodium  sulphantimonate  obtained  in 
the  last  preparation,  and  dilute  with  1  liter  of  cold  water.  Add  the 
sulphuric  acid  and  350  cc.  of  water  to  the  large  common  bottle. 
To  this  add  slowly,  and  with  constant  stirring,  the  solution  pre- 
pared above.  Fill  the  bottle  with  water  and  stir  thoroughly.  Let 
the  precipitate  settle,  draw  off  the  liquid,  and  wash  by  decantation 
until  the  wash  water  no  longer  gives,  with  barium  chloride,  the 
test  for  a  sulphate.  After  the  last  washing  let  settle  for  some  time, 
draw  off  as  much  as  possible  of  the  clear  liquid,  and  transfer  the 
slime  to  a  large  plain  filter  (Note  4  (c) ;  do  not  omit  to  reenf orce 
the  point  of  the  filter)  to  drain, for  12  hours  or  longer.  Without 
removing  the  pasty  antimony  sulphide  from  the  filter,  open  out  the 
latter  on  paper  towels,  and  leave  it  on  a  shelf  above  the  steam 
table  where  the  temperature  does  not  rise  above  50°.  When  the 
product  is  completely  dry,  detach  the  hardened  lumps  from  the 
paper  and  pulverize  them  in  a  mortar.  Put  up  the  product  in  a 
cork-stoppered  test  tube. 


METALLIC  ANTIMONY  277 

QUESTIONS 

1.  Write  equations  for  all  reactions  involved  in  the  preparation 
of  antimony  pentasulphide  from  stibnite. 

PREPARATION  54 
METALLIC  ANTIMONY 

This  metal  is  obtained  on  a  commercial  scale  both  by  reducing 
antimony  oxide  with  carbon  and  by  reducing  antimony  sulphide 
by  means  of  metallic  iron.  The  latter  method  possesses  the  ad- 
vantage that  antimony  sulphide,  a  natural  product,  is  used  directly 
and  does  not  need  to  be  first  converted  into  the  oxide.  The  iron 
sulphide  formed  by  this  method  is  fusible  and  forms  a  slag;  but 
the  slag  is  made  more  fusible  by  the  admixture  of  sodium  sulphate 
as  directed,  and  thus  the  globules  of  melted  antimony  are  allowed 
to  sink  more  easily  to  the  bottom  of  the  crucible  and  form  a  metallic 
regulus.  The  slag  furthermore  covers  the  surface  of  the  metal 
and  hinders  its  volatilization  and  oxidation. 

Materials:    stibnite,  Sb2S3,  112  grams.  =  }  F.W. 

iron  filings,  48  grams. 

anhydrous  sodium  sulphate,  12  grams. 

powdered  charcoal,  2  grams. 
Apparatus:  clay  crucible  of  250  cc.  capacity. 

gas  furnace. 

iron  stirrer,  use  old  file  gripped  with  furnace  tongs. 

• 

Procedure:  Mix  the  stibnite,  iron  filings,  sodium  sulphate,  and 
charcoal,  and  place  the  mixture  in  a  clay  crucible.  Cover  the 
crucible  tightly,  and  heat  it  in  the  gas  furnace  for  one  hour  at  a 
bright  red  heat.  The  temperature  should  not  be  high  enough  to 
volatilize  the  antimony,  which  would  in  that  case  escape  as  a  white 
smoke  consisting  of  antimony  oxide,  yet  the  slag  of  iron  sulphide 
must  be  completely  softened,  although  it  should  not  melt  to  a  thin 
liquid.  After  about  half  an  hour  test  the  conditions  by  removing 
the  cover  a  moment  and  stirring  the  slag  to  see  whether  it  is  in  the 
proper  semi-fluid  condition.  After  the  reaction  is  complete,  allow 
the  crucible  to  cool,  break  it  and  separate  the  regulus  of  antimony 
from  the  slag. 


278  ELEMENTS  OF  GROUP   V 

QUESTIONS 

1.  Warm  a  piece  of  metallic  antimony  with  hydrochloric  acid. 
Where  does  antimony  stand  in  the  electromotive  series? 

2.  Boil  J  gram  of  'powderecUantimony  in  a  small  casserole  with 
6-n  HNO3.     Describe  the  reaction  and  the  product  and  write 
equations. 

PREPARATION  55 
BISMUTH  BASIC  NITRATE  (BISMUTH  SUBNITRATE) 

Bismuth  is  the  most  strongly  metallic  element  of  the  fifth  group, 
yet  its  salts  in  aqueous  solution  undergo  partial  hydrolysis  very 
readily.  In  presence  of  a  considerable  amount  of  free  acid,  the 
Bi+++  ion  is  capable  of  existence  in  solution;  but  with  decreasing 
quantities  of  acid  the  tendency  to  hydrolyze  increases,  and  the 
basic  salt  of  bismuth,  which  is  only  slightly  soluble,  separates: 

3N03-  +  2H2O  ;=±  Bi(OH)2NO3  1  +  2H+  +  2NO3~. 


On  pouring  a  solution  of  bismuth  nitrate  into  a  considerable 
quantity  of  cold  water  the  basic  nitrate,  according  to  the  above 
formula,  is  precipitated.  This  salt,  however,  is  not  stable  in  con- 
tact with  a  solution  which  does  not  contain  nitric  acid  of  a  con- 
centration of  at  least  0.5-molal,  but  slowly  changes  over  into  some 
other  more  basic  nitrate,  and  if  washed  repeatedly  with  pure  water 
will  finally  go  over  completely  into  the  hydroxide: 

Bi(OH)2NO3  +  H20  ^±  Bi(OH)3  j  +  H+  +  NO3~. 

Under  the  conditions  in  the  following  procedure,  this  production 
of  a  more  basic  salt  will  occur  if  the  precipitate  is  allowed  to  stand 
in  contact  with  the  solution  for  a  considerable  time;  hence  the 
directions  to  filter  at  once. 

The  basic  nitrate  is  by  no  means  completely  insoluble  in  water, 
and  the  filtrate  contains  considerable  quantities  of  bismuth,  which 
can  be  conveniently  saved  as  oxide  by  precipitating  with  sodium 
carbonate. 

Materials:     crystallized  bismuth  nitrate,  Bi(NO3)3,  5H20,  42 

grams  =  TV  F.W. 
6-n  HNO3,  10  cc. 
solution. 


EXPERIMENTS  279 

Apparatus:  small  beaker. 

2-liter  common  bottle, 
suction  filter. 

Procedure:  Dissolve  without  heating  the  crystallized  bismuth 
nitrate,  in  10  cc.  of  6-n  HNOs  and  20  cc.  of  water.  Pour  this  into 
2  liters  of  cold  water  and  stir  thoroughly  for  a  few  minutes.  Let  the 
precipitate  settle  completely,  and  as  soon  as  this  has  occurred 
draw  off  and  save  the  supernatant  liquor;  drain  the  precipitate 
on  a  suction  filter,  and  wash  it  quickly  with  about  20  cc.  of  water. 
Dry  the  precipitate  at  the  steam  table,  and  preserve  it  as  a  powder 
in  a  cork-stoppered  test  tube. 

Bismuth  Oxide.  —  Combine  all  the  liquors  from  the  foregoing; 
add  sodium  carbonate  until  alkaline  to  litmus;  let  settle,  and  draw 
off  the  supernatant  liquor;  boil  the  remaining  suspension  after 
adding  to  it  about  20  grams  more  of  sodium  carbonate.  Then 
wash  the  precipitate  twice  by  decantation,  drain  on  a  suction  filter, 
and  wash  with  two  or  three  portions  of  water.  Dry  and  preserve 
this  product  in  a  cork-stoppered  test  tube. 

QUESTIONS 

1.  In  accordance  with  the  above  directions,  sodium  carbonate 
is  used  to  precipitate  bismuth  hydroxide.     Why  should  not  the 
precipitate  be  bismuth  carbonate? 

2.  If  this  precipitate  is  not  finally  boiled  with  an  excess  of 
sodium  carbonate,  it  is  likely  to  contain  a  certain  amount  of  basic 
nitrate.     Explain  why  this  should  be  so  and  why  the  boiling  will 
convert  it  completely  into  the  hydroxide. 

Experiments 

Review  in  Chap.  Ill  the  section  on  the  lonization  of  Polybasic 
Acids,  p.  102. 

in  Chap.  IV,  Preps.  7,  14,  and  Exps.  15,  25. 
in  Chap.  VIII,  Exps.  12,  13. 

1.  Oxidation  Products  of  the  Metals.  Treat  \  gram  each 
of  (a)  red  phosphorus,  (6)  powdered  arsenic,  (c)  powdered 
antimony,  and  (d)  powdered  bismuth  with  excess  of  6-n 
HNO3  (10-15  cc.)  and  note  that  red  gases  are  evolved  in 
each  case. 

(a)  A  clear  solution  results.     Evaporate  this  until  excess 


280  ELEMENTS  OF  GROUP  V 

of  volatile  nitric  acid  is  expelled  but  do  not  exceed  a  tem- 
perature of  180°.  A  sirupy  liquid  is  left  which  dissolves  in 
water  to  give  an  acid  solution. 

(6)  A  clear  solution  is  left.  Evaporate  this  carefully  to 
dryness.  A  white  solid  isjFeft  which  dissolves  in  water  to  give 
an  acid  solution. 

(c)  The  metal  disappears  and  a  white  powdery  solid  remains 
in  the  liquid.     This  solid  is  insoluble  in '-nitric  acid  or  water. 

(d)  A  clear  solution  results  out  of  which  after  concentrating 
and  cooling,  clear  crystals  of  a  salt  separate.     These  crystals 
dissolve  in  a  little  water  acidified  with  HNOs  to  give  a  clear 
solution,  but  a  white  precipitate  is  formed  if  the  solution  is 
diluted  with  a  large  amount  of  water. 

Phosphorus,  arsenic  and  antimony  are  oxidized  by  nitric  acid  to 
hydrated  forms  of  the  pentoxides,  giving  respectively:  phosphoric 
acid,  H3PO4,  which  is  very  soluble  in  water  and  a  fairly  strong  acid; 
arsenic  acid,  HsAsO^  which  can  be  dehydrated  to  the  oxide  As2O5 
which  will  dissolve  in  water  again  to  form  the  acid;  meta-anti- 
monic  acid,  which  is  a  very  weak  and  insoluble  acid.  Bismuth  is 
oxidized  only  to  the  trivalent  condition;  Bi2O3  is  basic  and  forms 
the  salt  Bi(NOs)3  with  the  excess  of  HNOs.  This  salt  hydrolyzes 
very  easily  to  an  insoluble  salt  and  with  a  large  amount  of  water  it 
hydrolyzes  completely  to  Bi(OH)3. 

2.  Sulphides  and  Thio-Salts.  Pass  hydrogen  sulphide 
into  hot  dilute  solutions  of  arsenic,  antimony,  and  bismuth 
trichlorides  in  separate  test  tubes.  Note  that  yellow,  orange, 
and  black  precipitates  respectively  are  formed.  Let  the 
precipitates  settle  to  the  bottom  of  the  tubes,  pour  off  the 
liquid  and  treat  the  solid  with  ammonium  polysulphide 
((NH4)2S  solution  in  which  sulphur  is  dissolved)  in  each 
case.  The  yellow  and  orange  precipitates  dissolve,  the  black 
one  does  not.  To  the  two  solutions  add  6-n  HC1  in  excess 
and  note  that  yellow  and  orange  precipitates  respectively  are 
again  thrown  out. 

Review  the  discussion  of  thio-  or  sulpho-salts  under  Prep. 
52.  The  trisulphides  As2S3  and  Sb2S3  are  oxidized  by  the  free 
sulphur  to  the  pentasulphides,  As2S5  and  Sb2S5,  which  react  with 
the  (NH4)2S  to  form  the  soluble  thio-salts,  NasAsS4  and  Na3SbS4. 


EXPERIMENTS  281 

The  addition  of  HC1  displaces  the  weak  thio-acids,  HsAsS4  and 
H3SbS4,  which  are  too  unstable  to  exist  alone  and  decompose  into 
H2S  and  the  respective  pentasulphides.  The  color  of  the  penta- 
sulphides  seems  to  be  identical  to  that  of  the  trisulphides.  Bis- 
muth shows  no  tendency  to  form  a  thio-salt. 

3.  Reducing    Action    of    Phosphorous    Acid;     Non-Oxidizing 
Property  of  Phosphoric  Acid.     Review  the  discussion  of  Prep.  49, 
and  the  test  employed  for  phosphorous  acid.     Phosphorous  acid 
reduces  silver  nitrate  to  metallic  silver,  it  itself  being  oxidized  to 
phosphoric  acid. 

Recall  that  in  Exp.  15,  Chap.  IV,  phosphoric  acid  did  not 
oxidize  hydrogen  bromide  or  hydrogen  iodide,  which  are  excep- 
tionally strong  reducing  agents. 

4.  Arsenious  and  Arsenic  Acids.     Review  Note  1  and  Exp. 
3  under  Prep.  50.     Arsenious  acid  reduces  iodine  to  hydriodic  acid 
in  a  solution  containing  sodium  bicarbonate,  the  latter  neutral- 
izing the  acid  produced : 

H3AsO3  +  Ia  +  H2O  <=>  H3As04  +  2HI; 

but  in  an  acid  solution  the  reaction  goes  in  the  opposite  direction, 

the  hydrogen  iodide  being  oxidized. 

The  reaction  progressing  to  the  right  according  to  the  equation 

produces  H+  ions.     It  is  natural  that  the  removal  of  H+  ions 

favors  the  reaction  and  their  presence  reverses  it. 

5.  Reduction  of  Bismuth  Salts,  (a)  Suspend  about  2  milli- 
grams of  bismuth  subnitrate  in  10  cc.  water,  add  5  cc.  6-n 
NaOH  boil  and  note  that  the  white  suspension  does  not  change 
color.  Add  a  few  drops  of  grape  sugar  solution,  continue  to 
boil  and  note  a  black  precipitate. 

(6)  Again  suspend  about  2  milligrams  bismuth  subnitrate  in 
10  cc.  boiling  water.  Pour  through  a  filter  leaving  the  white 
Bi(OH)3,  to  which  the  salt  is  hydrolyzed,  on  the  paper.  Make 
a  sodium  stannite  solution  by  adding  drop  by  drop  6-n  NaOH 
to  2  cc.  of  SnCl2  solution  with  constant  shaking  and  cooling 
until  the  precipitate  formed  at  first  is  redissolved.  Pour  this 
solution  over  the  filter  paper  and  note  the  intense  black  color. 

Bismuth  stands  low  in  the  electromotive  series  and  its  salts 
are  easily  reduced  to  the  metal  which  in  the  finely  divided  state  is 
intensely  black. 

2Bi(OH)3  +  3Na2Sn02  ->  2Bi  j  +  3Na2SnO3  +  3H2O. 


282  ELEMENTS  OF  GROUP  V 

Sugar  may  be  tested  for  in  urine  by  an  adaptation  of  the  reaction 
in  (a) ;  its  presence  indicates  diabetes. 

6.  Bismuth  in  a  Higher  State  of  Oxidation.  Fusion  of  bismuth 
salts  with  sodium  hydroxide  and  an  oxidizing  agent  yields  a 
material  which  has  beeri  called  podium  bismuthate,  from  the  hypo- 
thetical oxide  Bi2O5.  Such  a  compound  has  never  been  obtained 
pure.  If  the  melt  is  extracted  with  water,  the  salt  hydrolyzes 
completely  and  analysis  of  the  brown  residue- gives  a  composition 
approximating  the  formula  BiO2,  rather  than  Bi2O5. 

Add  1  drop  of  Mn(NO3)2  or  MnSO4  solution  (but  not 
MnCl2)  to  5  cc.  of  cold  6-n  HNO3.  Add  about  -^  gram  solid 
bismuth  dioxide,  agitate,  let  the  brown  solid  settle  and  note 
the  deep  red  permanganate  color  of  the  solution. 

Bismuth  in  a  higher  state  of  oxidation  than  that  corresponding 
to  the  oxide  Bi203  must  be  a  very  strong  oxidizing  agent  if  it  can 
oxidize  a  manganese  compound  to  permanganic  acid. 

Mn(NO3)2  +  13HN03  +  5Bi02  ->  5Bi(N03)3  +  HMn04  +  6H20 

GENERAL  QUESTIONS  X 
ELEMENTS  OF  GROUP  V 

1.  State  in  each  case  whether  the  nitrate  or  sulphate  of  tri- 
valent  phosphorus,  arsenic,  antimony,  or  bismuth  can  be  prepared, 
and  if  so  whether  it  can  be  dissolved  in  water  without  suffering 
complete  hydrolysis.     How  does  the  basic  nature  of  the  trioxide 
change  in  the  series,  phosphorus  to  bismuth?     Can  nitrates  or 
sulphates  of  any  of  these  elements  in  their  pentavalent  condition 
be  prepared?     For  any  one  of  the.  elements,  which  is  the  more 
strongly  basic  in  nature,  the  trioxide  or  the  pentoxide?     Which 
is  the  more  strongly  acidic?     Give  the  symbol  of  the  most  common 
acid,  if  one  exists,  which  is  derived  from  the  pentoxide  of  each  of 
these  elements.     How  does  the  acidic  nature  of  the  pentoxide 
change  in  passing  from  nitrogen  to  bismuth? 

2.  Name  the  simplest  hydrogen  compounds  of  nitrogen,  phos- 
phorus, arsenic,  and  bismuth.     Compare  the  stability  of  these 
hydrides  when  heated.     Compare  any  ability  they  may  possess  to 
unite  with  water  to  form  bases,  and  with  acids  to  form  salts. 

3.  Write  the  equations  for  the  reaction  of  nitric  acid  with 
phosphorus,  arsenic,  antimony,  and  bismuth,  respectively. 


GENERAL  QUESTIONS  X  283 

4.  How  do  the  trisulphides  of  arsenic  and  antimony  behave 
when  treated  with  a  solution  of  ammonium  sulphide,  (NH4)2S? 
With  a  solution  of  ammonium  polysulphide,  (NH4)2S.SX?  How 
does  the  solution  obtained  with  ammonium  polysulphide  behave 
when  it  is  acidified?  Give  equations  for  all  the  reactions. 

What  is  the  relation  between  sulpho-  and  oxy-acids?  Show, 
for  example,  how  sodium  sulpharsenate  is  derived  from  two  simple 
sulphides,  and  sodium  arsenate  from  the  corresponding  oxides. 


v      CHAPTER  XI 

*     •       j*- 

HEAVY  METALS  OF  CROUPS  VI,  VII,  AND  VIII 
OF  THE  PERIODIC   SYSTEM 

An  inspection  of  the  Periodic  Table  of  the  elements  shows  that 
chromium,  manganese,  iron,  cobalt,  and  nickel,  and  following 
these  copper  and  zinc,  come  in  the  middle  portion  of  the  long 
period  that  begins  with  potassium  and  ends  with  bromine.  The 
seven  elements  mentioned  possess  high  specific  gravities  and  all 
come  under  the  classification  of  heavy  metals.  In  certain  of  their 
compounds  they  are  extremely  similar  to  one  another;  in  other 
of  their  properties  they  are  very  dissimilar  and  exhibit  the  chemical 
characteristics  of  the  respective  groups  to  which  they  belong. 

The  heavy  metals  occupying  a  corresponding  position  in  the 
middle  of  the  next  long  period  are  molybdenum,  an  unknown 
element  which  should  come  below  manganese,  ruthenium,  rhodium, 
palladium,  silver,  and  cadmium.  In  the  middle  of  the  next  long 
period  come  tungsten,  another  unknown  element  which  should 
resemble  manganese,  osmium,  iridium,  platinum,  gold,  and 
mercury.  In  the  last  long  period,  of  which  there  is  at  best  only  a 
fragmentary  indication,  the  only  representative  of  this  class  of 
heavy  metals  is  uranium. 

In  the  sixth  group,  chromium,  molybdenum,  tungsten,  and 
uranium  constitute  Family  A.  In  their  trioxides  they  show  the 
characteristic  valence  of  the  sixth  group  and  resemble  in  properties 
the  non-metals  of  Family  B,  of  which  sulphur  is  the  type.  In  their 
lower  oxides  they  show  none  of  the  group  characteristics,  but  show 
the  general  base-forming  properties  of  the  heavy  metals. 

In  the  Group  VII,  manganese,  the  only  known  representative  of 
Family  A,  resembles  the  halogens  in  its  heptoxide,  M^Oy;  in  its 
lower  oxides  it  shows  no  resemblance  to  the  halogens,  but  does 
show  properties  similar  to  those  of  other  heavy  metals  when  they 
are  in  the  same  state  of  oxidation;  in  its  lowest  oxide,  MnO,  it  is 
distinctly  a  base-forming  element. 

In  the  Group  VIII,  each  position  instead  of  being  filled  by  a 
single  element  is  occupied  by  a  group  of  three  elements.  Thus 

284 


POTASSIUM   BICHROMATE  FROM   CHROMITE  285 

there  appear  in  triads:  iron,  cobalt,  and  nickel;  ruthenium, 
rhodium,  and  palladium;  and  osmium,  iridium,  and  platinum. 
In  this  group  there  is  no  subdivision  into  families,  but  all  of  the 
members  are  heavy  metals.  The  Zero  Group,  which  comprises 
the  inert  gases,  helium,  neon,  argon,  krypton,  and  xenon,  may  be 
regarded  as  bearing  the  same  relation  to  Group  VIII  as  Family  B 
of  an  ordinary  group  bears  to  Family  A.  If  this  view  is  a  correct 
one,  the  divergence  in  properties  between  the  families  is  in  this 
case  at  a  maximum. 

Of  the  heavy  metals  discussed  above,  the  ones  that  are  of 
most  frequent  occurrence  and  that  are  to  receive  detailed  treat- 
ment in  this  chapter  are  chromium,  manganese,  and  iron. 

PREPARATION  56 
POTASSIUM  BICHROMATE  FROM  CHROMITE 

The  most  important  source  of  chromium  is  the  mineral  chromite, 
FeO.Cr2O3  or  Fe(CrO2)2.  This  substance,  as  indicated  by  the 
formula,  may  be  regarded  as  a  compound  of  ferrous  oxide  and 
chromic  oxide,  or  as  a  salt,  chromite  of  iron,  in  which  ferrous  oxide 
is  the  basic  constituent  and  chromic  oxide  the  acidic.  Chromite 
is  a  difficult  material  to  decompose,  and  the  ordinary  method  by 
which  this  is  done  is  treatment  at  a  high  temperature  with  an 
alkali  and  an  oxidizing  agent.  The  iron  is  thereby  converted  to 
the  ferric  condition  (Fe2O3),  and  the  chromium  is  oxidized  to  the 
hexavalent  condition  (CrO3),  at  the  same  time  combining  with  the 
alkali  to  form  a  chromate  (K2CrO4). 

In  the  commercial  method  for  manufacturing  chromates, 
atmospheric  oxygen  is  the  oxidizing  agent.  The  chromite  is  mixed 
with  potassium  carbonate  and  calcium  carbonate,  the  latter  to 
give  porosity,  and  then  heated  for  a  considerable  time  in  a  furnace 
with  free  access  of  air.  The  chromium  trioxide,  CrOs,  produced 
by  the  oxidation  reacts  with  the  potassium  carbonate,  displacing 
carbon  dioxide  and  giving  potassium  chromate,  K2CrO4.  After 
cooling,  the  contents  of  the  furnace  are  treated  with  a  solution 
of  sodium  sulphate;  the  potassium  chromate  dissolves,  the  iron 
oxide  is  insoluble,  and  the  calcium  oxide  (from  heating  the  carbon- 
ate) reacts  with  the  sodium  sulphate  to  form  insoluble  calcium 
sulphate.  From  the  solution  potassium  chromate  could  be 
crystallized  but  for  the  fact  that  it  is  very  soluble  in  water  and 


286       <  HEAVY  METALS  OF  GROUPS  VI,   VII,   AND  VIII 

could  not  be  separated  from  the  other  salts.  Potassium  dichro- 
mate,  however,  is  much  less  soluble,  and  it  may  be  formed  by  add- 
ing sulphuric  acid, 

.  2K2CrO4  +  H2SO4  =  K2SO4  +  K2Cr2O7  +  H2O, 


and  obtained  pure  by  crystallization. 

On  account  of  the  difficulty  of  carrying  out  the  above  process 
on  a  laboratory  scale,  the  following  less  economical  procedure  is 
given  in  which  the  mineral  is  heated  with  considerably  more  than 
the  quantity  of  potassium  carbonate  theoretically  required,  in 
order  to  give  a  more  liquid  melt,  and  with  potassium  nitrate  for 
the  oxidizing  agent.  The  solution  obtained  by  extracting  this 
melt  with  water  contains  so  much  potassium  carbonate  that  it 
would  be  very  difficult  to  separate  the  also  very  soluble  .potassium 
chromate  from  it  by  crystallization.  If  acetic  acid  is  added  until 
the  solution  reacts  slightly  acid,  the  potassium  carbonate  is  con- 
verted into  the  very  soluble  acetate,  and  the  chromate  is  changed 
to  the  only  moderately  soluble  potassium  dichromate  which, 
especially  in  the  presence  of  the  large  amount  of  the  other  salt 
with  the  K+-ion  in  common,  can  be  crystallized  out. 


A  saturated  solution  contains  for  each  100  grams  of  water  the  given 
number  of  grams  of  the  anhydrous  salt. 


Temperature  

0° 

10° 

20° 

30° 

40° 

50° 

70° 

100° 

K2CrO4         

59 

61 

63 

65 

67 

69 

73 

79 

K2Cr2O7  

5 

7 

12 

20 

26 

35 

55 

88 

Materials:  finely  powdered  chromite,  Fe(CrO2)2,  37  grams  = 
i  F.W.;  unless  very  pure  mineral  is  available 
better  results  will  be  obtained  by  using  25  grams 
of  artificially  prepared  chromic  oxide,  Cr2Oa. 

anhydrous  potassium  carbonate,  K2COs,  92  grams. 

potassium  nitrate,  28  grams. 

glacial  acetic  acid. 
Apparatus:  cast  iron  crucible  of  350  cc.  capacity. 

gas  furnace. 

8-inch  evaporating  dish. 

Procedure:  Mix  the  chromite,  potassium  carbonate,  and  potas- 
sium nitrate  in  the  cast  iron  crucible  which  must  be  on  no  account 


POTASSIUM  CHROMATE  FROM  POTASSIUM  BICHROMATE     287 

more  than  two-thirds  filled  with  the  dry  mixture.  Heat  in  the 
gas  furnace  to  a  yellow  heat  until  the  melted  charge  has  ceased  to 
effervesce,  but  use  great  caution  not  to  melt  the  crucible  nor  to 
let  the  melted  charge  overflow.  Pour  the  molten  mass  out  onto 
a  dry  iron  plate.  When  cool  crack  it  up  and  dissolve  it,  together 
with  what  still  adheres  to  the  crucible,  in  boiling  water.  Filter 
the  solution,  and  extract  the  residue  with  a  little  more  boiling 
water  and  pour  through  the  same  filter.  Add  glacial  acetic  acid 
(cautiously)  to  the  filtrate  until  it  has  become  acid.  Boil  down  the 
solution  to  300  cc.,  or  to  even  a  less  volume  if  no  solid  salt  begins 
to  separate.  Add  25  cc.  more  of  glacial  acetic  acid,  let  stand  for 
some  time,  and  finally  cool  to  0°  before  separating  the  crystal  meal 
of  potassium  dichromate  from  the  mother  liquor.  Purify  the 
product  by  recrystallization,  and  put  it  up  in  a  cork-stoppered 
test  tube. 

QUESTIONS 

1.  Name  at  least  three  oxidizing  agents  which  might  have  been 
used  instead  of  potassium  nitrate  in  this  preparation,  and  write 
equations. 

•  2.  To  5  cc.  of  a  chromic  salt  solution  (CrCl3  or  Cr2(S04)3)  add 
NaOH  in  excess;  cool,  add  about  1  gram  of  sodium  peroxide, 
agitate  for  a  few  minutes  and  then  boil  until  effervescence  ceases. 
Describe  observations  and  write  equations. 

3.  To  a  solution  of  potassium  dichromate  add  K2C03  until  no 
more  effervescence  takes  place.     Explain  the  effervescence  and 
the  change  in  color.     Write  equation. 

4.  To  the  solution  from  Exp.  3  add  6-n  H2SO4,  observe  and 
explain  as  before.     Explain  fully  the  difference  between  chromates 
and  dichromates. 

5.  Show  that  potassium  acid  sulphate,  KHS04,  and  potassium 
dichromate,  are  very  similar,  differing  mainly  in  the  degree  of 
hydration. 

PKEPARATION  57 

POTASSIUM  CHROMATE  FROM  POTASSIUM  BICHROMATE 

Dissolve  49  grams  of  potassium  dichromate  in  water  and  add 
the  calculated  amount  of  potassium  carbonate  dissolved  in  water. 
Crystallize  the  product  from  the  solution  (see  solubility  table  on 
page  286). 

Answer  the  questions  given  under  Potassium  Dichromate. 


288       HEAVY  METALS  OF  GROUPS  VI,   VII,   AND  VIII 

PREPARATION  58 
CHROMIC  ANHYDRIDE,  CrO3 

When  a  chromate  or  a  dichromate  is  treated  with  a  strong  acid, 
chromic  acid  is  formed-  in  the  Solution.  The  affinity  of  chromic 
anhydride  for  water  is  far  less  than  that  of  sulphuric  anhydride  for 
water;  and  chromic  acid,  therefore,  instead  of  existing  in  solution 
entirely  in  the  form  H2CrO4,  is  broken  down  to  a  great  extent  into 
H2Cr207  (i.e.,  H20.2CrO3)  and  even  to  CrO3.  Especially  in  the 
presence  of  a  large  amount  of  sulphuric  acid,  the  last  form  is  pro- 
duced so  freely  that  it  crystallizes  out  in  the  shape  of  red  needles. 

Materials:     sodium  dichromate,  Na2Cr2O7.2H2O,  100  grams  = 
1  F.W. 

36-n  H2SO4,  400  cc. 
Apparatus:  8-inch  evaporating  dish. 

glass  plate  to  cover  the  8-inch  dish. 

suction  filter  with  glass  marble. 

unglazed  porcelain  plate. 

glass-stoppered  sample  bottle. 

Procedure:  Dissolve  the  100  grams  of  sodium  dichromate  in  250 
cc.  of  water  and  filter  from  any  sediment.  Add  rather  slowly  with 
constant  stirring  about  half  of  the  concentrated  sulphuric  acid 
until  a  slight  permanent  precipitate  of  Cr03  is  formed.  Let  the 
mixture  cool  for  half  an  hour  or  longer,  then  add  slowly,  while 
stirring,  the  rest  of  the  sulphuric  acid.  Let  the  mixture  stand 
over  night  covered  with  a  glass  plate  in  order  that  the  crystal 
meal  may  become  somewhat  coarser.  In  such  a  crystal  meal 
standing  in  its  saturated  solution,  the  smaller  grains  dissolve  and 
their  material  deposits  out  on  the  larger  crystals.  But  even  now 
the  crystal  meal  will  be  rather  fine  and  it  will  at  first  run  through 
the  filter;  if,  however,  while  waiting,  the  mixture  is  heated  with 
stirring  to  100°  and  allowed  to  cool  slowly,  and  this  process  is 
repeated  once  or  twice,  a  more  satisfactory  product  will  be  ob- 
tained. To  collect  the  crystals,  use  a  suction  filter,  but  place  a 
smal  glass  marble  in  the  funnel  instead  of  the  usual  plate  and  paper. 
If  the  red  crystals  at  first  run  past  the  sides  of  the  marble,  pour  the 
liquid  in  the  bottle  repeatedly  back  onto  the  filter  until  finally 
the  filtrate  runs  clear  (see  last  sentence  of  Note  3  on  page  5). 
After  draining  the  crystals  completely  and  pressing  the  surface 


AMMONIUM  BICHROMATE  289 

with  a  glass  spatula,  stop  the  suction  and  pour  15  cc.  of  16-n  HNO3 
so  as  to  wash  down  the  sides  of  the  funnel  and  cover  the  surface 
of  the  product.  Stir  up  the  product  with  this  washing  fluid  for  a 
depth  of  about  J  inch.  Suck  dry  and  repeat  the  operation  twice 
with  10  cc.  of  nitric  acid  each  time.  Finally  transfer  the  product 
to  an  unglazed  porcelain  plate,  place  the  latter  on  an  iron  ring  and 
heat  it  by  playing  under  it  the  burner  held  in  one  hand  while  with 
the  other  hand  the  crystals  are  continually  stirred.  Continue 
this  operation,  being  very  careful  not  to  overheat,  until  nitric  acid 
vapors  cease  to  be  given  off.  Transfer  the  product  at  once  to  a 
dry,  previously  weighed,  glass-stoppered  bottle. 

QUESTIONS 

1.  Dissolve  \  gram  of  chromic  anhydride  in  a  few  drops  of  water. 
What  is  the  color  of  the  solution?     Dilute  with  200  cc.  water. 
What  is  the  color?     Write  ionic  equations  showing  the  equilibrium 
condition  among  the  different  acids  of  chromium  and  their  ions 
in  the  solution.     Show  that  according  to  the  law  of  molecular  con- 
centration the  proportion  of  the  yellow  to  the  red  components 
should  increase  as  the  solution  is  diluted  with  water. 

2.  Heat  a  little  chromic  anhydride  strongly  on  a  bit  of  porcelain. 
Pulverize  the  residue  in  a  white  mortar  so  as  to  better  observe  its 
color.     What  is  the  residue? 

PREPARATION  59 
AMMONIUM  DICHROMATE,  (NH4)2Cr207 

This  salt  is  prepared  by  crystallization  from  a  solution  of  chromic 
anhydride  to  which  the  calculated  amount  of  NH4OH  is  added. 
It  is  of  particular  interest  because  one  radical  is  an  oxidizing  agent 
and  the  other  a  reducing  agent,  in  consequence  of  which  if  the  dry 
salt  is  heated  a  little,  a  self-propagating  reaction  which  is  rather 
spectacular  takes  place. 

Materials:     chromic  anhydride,  CrOs,  50  grams  =  J  F.W. 

14-n  NH4OH,  36  cc. 
Apparatus:  300  cc.  flask. 

4-inch  crystallizing  dish. 

Procedure:  Dissolve  the  chromic  anhydride  in  100  cc.  water  in 
a  flask.  Add  the  NH4OH  cautiously  so  as  not  to  volatilize  am- 


290       HEAVY  METALS  OF  GROUPS  VI,   VII,   AND  VIII 

monia,  and  leave  the  solution  to  crystallize  in  a  4-inch  dish  After 
getting  one  crop  of  crystals  evaporate  the  mother  liquor  somewhat 
and  obtain  more  crystals.  Dry  the  product  on  paper  towels  at 
room  temperature  and  preserve  it  in  a  2-ounce  cork-stoppered 
bottle.  p. 

QUESTIONS 

1.  Heap  up  10  grams  of  ammonium  dichromate  in  a  small 
mound  on  a  porcelain  plate,  and  apply  the  flame  to  the  top  of  the 
mound    until    a    reaction    starts.     Write    the    equation.     What 
element  is  oxidized  and  what  one  reduced?     Show  that  the  alge- 
braic sum  of  the  valence  changes  is  zero. 

2.  Suggest    a    method    of    preparing    ammonium    chromate, 
(NH4)2CrO4. 

PREPARATION  60 
CHROMIC  ALUM 

The  preparation  of  potassium  dichromate  illustrated  how 
chromic  oxide,  Cr20s,  as  it  exists  in  nature  as  a  constituent  of  the 
mineral  chromite,  can  be  oxidized  to  a  chromate  in  which  chromium 
exists  as  CrOs.  For  the  preparation  of  chromic  alum,  K2S04.- 
Cr2(S04)3.24H20,  it  might  seem  as  if  chromite  should  yield  chromic 
sulphate  directly  on  treatment  with  sulphuric  acid.  This  is, 
however,  impossible,  because  the  natural  material  is  very  resistant 
to  the  action  of  acids.  It  yields  only  to  the  action  of  alkaline 
oxidizing  agents,  which  convert  it  into  a  chromate.  Therefore 
potassium,  or  sodium,  dichromates  are  always  the  products  made 
directly  from  the  mineral,  and  these  serve  as  the  materials  from 
which  other  compounds  of  chromium  are  prepared.  To  make 
chromic  alum  from  potassium  dichromate  it  is  necessary  to  reduce 
the  chromium  to  the  state  of  oxidation  in  which  it  originally  existed 
in  the  mineral,  and  to  add  sufficient  sulphuric  acid  to  form  the 
sulphates  of  potassium  and  chromium.  Alcohol  may  be  used  as 
the  reducing  agent,  it  being  itself  oxidized  to  aldehyde,  C2H4O, 
a  substance  whose  presence  is  made  very  evident  by  its  pene- 
trating odor. 

Chromic  alum  is  isomorphous  with  common  alum  and  can  easily 
be  obtained  in  large  and  beautiful  deep  purple  crystals.  Care 
must,  however,  be  exercised  not  to  allow  the  temperature  of  its 


CHROMIC  ALUM  291 

solution  to  rise  above  50°  during  the  preparation,  for  when  heated 
beyond  this  point  it  undergoes  a  change  into  a  green  non-crystal- 
lizable  substance.  This  green  substance  is  not  stable  at  the 
ordinary  temperature,  and  after  cooling  it  will  change  slowly  back 
into  the  ordinary  crystallizable  chromic  alum;  but  so  slowly, 
however,  that  if  once  it  is  formed  the  preparation  is  practically 
spoiled. 

At  25°,  24  grams  of  KaSO^C^SO^s^HaO  will  dissolve  in 
100  grams  of  water,  and  the  solubility  increases  very  rapidly  with 
the  temperature. 


Materials:     potassium  dichromate,  K^C^Oy,  98  grams  =  J  F.W. 

36-n  H2SO4,  76  cc. 

alcohol,  63  cc. 
Apparatus:  8-inch  evaporating  dish. 

thermometer. 

Procedure:  Pulverize  98  grams  of  potassium  dichromate,  and 
cover  it  in  an  8-inch  evaporating  dish  with  400  cc.  of  water.  Add 
76  cc.  of  concentrated  sulphuric  acid,  and  stir  until  the  salt  is  all 
dissolved.  Adding  the  sulphuric  acid  should  produce  enough  heat 
to  dissolve  the  dichromate,  but  if  it  is  necessary  heat  the  mixture 
a  little  more.  Be  sure  that  the  last  trace  of  solid  is  dissolved. 
Allow  the  solution  to  cool  to  40°;  then  add  alcohol,  a  drop  at  a 
time,  while  stirring  constantly  with  the  stem  of  a  thermometer 
until  the  temperature  commences  to  rise.  Then  place  the  dish 
in  a  pan  of  ice  and  water  and  add  alcohol,  63  cc.  in  all,  at  first  very 
slowly,  endeavoring  to  keep  the  temperature  between  35°  and  45°, 
and  finally  more  rapidly.  Keep  the  temperature  at  all  times 
below  50°,  and  if  it  should  start  to  rise  suddenly,  due  to  too  large 
an  addition  of  alcohol,  and  get  as  high  as  50°,  drop  a  piece  of  ice 
directly  into  the  solution.  Finally,  let  the  solution  cool  completely 
in  the  bath  of  ice  water,  or,  still  better,  let  it  stand  over  night. 
Collect  the  crystal  meal  on  a  suction  filter  and  free  it  from  liquid. 
Recrystallize  so  as  to  obtain  large,  well-shaped  crystals,  following 
a  similar  procedure  and  observing  the  same  precautions  as  with 
common  alum  (see  page  181).  A  saturated  solution  of  this  salt 
should  be  prepared  at  35°  (Note  7,  p.  12).  After  freeing  it  of 
any  undissolved  particles  of  the  crystal  meal,  warm  it  to  45°,  and 
set  it  to  crystallize,  with  the  addition  of  about  ten  very  small 


292        HEAVY  METALS  OF  GROUPS  VI,   VII,   AND  VIII 

crystals  to  serve  as  nuclei.  Dry  the  crystals  by  leaving  them 
wrapped  in  paper  towels  over  night  (Note  9(6),  p.  15),  and  then 
stopper  them  at  once  in  a  bottle,  since  they  are  efflorescent. 

QUESTIONS 

1  Write  equations  for  the  following  steps  in  the  reaction  of 
potassium  dichromate  in  acid  solution  with  alcohol,  (a)  Resolve 
the  salt  into  its  basic  and  acidic  anhydrides,  (b)  Let  the  acid 
anhydride,  which  is  the  higher  oxide  of  chromium,  be  reduced  by 
the  alcohol  to  the  lower  oxide,  Cr2O3.  (c)  Let  the  sulphuric  acid 
form  salts  with  the  basic  anhydride  and  the  lower  oxide  of  chro- 
mium, which  is  also  a  basic  oxide.  Add  the  steps  together  to  give 
the  complete  equation. 

2.  Sulphur  dioxide  might  serve  as  the  reducing  agent.     Write 
equations  for  the  reaction  in  steps  and  add  the  equations. 

3.  Describe  the  effect   observed  when  hydrogen   sulphide  is 
passed  into  a  hot  acidified  solution  of  K2Cr207.     Write  equation 
in  one  line  marking  with  Roman  numerals  the  valences  of  sulphur 
and  chromium  and  showing  that  the  algebraic  sum  of  the  valence 
changes  is  zero.     (See  Exps.  10  and  23,  Chap.  IV.) 

PREPARATION  61 
CHROMIUM  METAL  BY  THE  GOLDSCHMIDT  PROCESS 

The  readiest  method  of  obtaining  the  metal  chromium  from  its 
oxide,  and  one  which  yields  it  in  a  high  state  of  purity,  is  the  so- 
called  Goldschmidt,  or  alumino-thermic  process,  in  which  use  is 
made  of  metallic  aluminum  as  the  reducing  agent  according  to  the 
reaction, 

2A1  +  Cr203  =  A1203  +  2Cr. 

The  heat  produced  by  the  oxidation  of  aluminum  is  so  great  that 
it  is  sufficient  to  effect  the  decomposition  of  the  chromic  oxide  with 
still  enough  surplus  heat  to  produce  a  temperature  high  enough 
to  melt  the  metallic  chromium.  It  is  evident  that  before  this 
reaction  can  be  made  to  progress  spontaneously  a  sufficient 
temperature  must  be  developed  to  decompose  the  chromium  oxide. 
This  necessary  temperature  is  a  good  deal  higher  than  that  of  a 
flame  or  of  a  common  furnace,  but  can  be  obtained  by  use  of  the 


CHROMIUM   METAL  293 

fuse  powder  described  below.  When  once  started  in  this  way  the 
reaction  itself  produces  a  temperature  high  enough  to  insure  its 
continuance. 

Carried  out  on  the  small  scale  of  a  laboratory  preparation,  the 
heat  produced  is  not  quite  sufficient  to  melt  the  metal  and  slag 
so  thoroughly  that  the  metal  can  settle  out  to  form  a  compact 
regulus  at  the  bottom  of  the  crucible.  By  adding  a  small  amount 
of  potassium  dichromate  to  the  charge,  however,  the  reaction 
becomes  more  energetic,  owing  to  the  more  available  supply  of 
oxygen. 

Materials:     chromic  oxide,  Cr203,  210  grams. 

potassium  dichromate,  60  grams. 

granulated  aluminum,  96  grams. 

barium  peroxide,  20  grams. 
Apparatus:  clay  crucible  of  600  cc.  capacity. 

gas  furnace. 

iron  pan. 

pail  of  sand. 

Procedure:  Heat  the  chromic  oxide  in  the  crucible  in  the  gas 
furnace  for  40  minutes  or  longer.  Melt  the  potassium  dichromate 
in  a  clean  iron  pan  and  pulverize  it  in  a  mortar  after  it  has  solidified. 
Mix  the  chromic  oxide,  potassium  dichromate,  and  granulated 
aluminum  thoroughly  in  a  mortar.  Make  a  fuse  powder  with 
2  grams  of  granulated  aluminum  and  20  grams  barium  peroxide. 
Take  half  of  the  fuse  powder  and  mix  it  with  twice  its  bulk  of 
the  main  charge.  Hold  a  rather  wide  test  tube  in  the  middle  of 
the  crucible,  pack  the  charge  around  it  and  withdraw  it  carefully 
leaving  a  deep  hole  in  the  middle.  Carefully  pour  the  mixture 
of  fuse  powder  and  charge  into  the  bottom  of  the  hole;  pour  the 
fuse  powder  on  top  of  it  in  the  hole;  and  insert  a  strip  of  mag- 
nesium ribbon  into  the  fuse  powder.  Imbed  the  crucible  in  a  pail 
of  hot  sand  and  place  the  latter  under  the  hood  at  a  distance  from 
any  woodwork.  Start  the  reaction  by  igniting  the  end  of  the  mag- 
nesium ribbon  with  a  gas  flame.  It  is  advisable  for  the  operator 
to  wear  colored  glasses  while  watching  the  reaction,  and  to  keep 
at  a  little  distance  to  be  out  of  the  way  of  flying  sparks.  When 
the  crucible  has  cooled,  break  it  and  separate  the  regulus  of  metal- 
lic chromium  from  the  slag  of  fused  aluminum  oxide. 


294       HEAVY  METALS  OF  GROUPS  VI,   VII,   AND  VIII 


1.  What  is  approximately  the  position  of  chromium  in  the 
electromotive  series?     What  bearing  does  this  have  upon  the 
question  of  reducing  chromic  £xide? 

2.  What  metals  can  be  used'in  place  of  aluminum  in  the  alumino- 
thermic  process? 

3.  What  other  metals  than  chromium  can  be  advantageously 
prepared  by  this  process,  and  why? 

PREPARATION  62 
MANGANESE  CHLORIDE  FROM  WASTE  MANGANESE  LIQUORS 

The  waste  liquors  left  after  the  generation  of  chlorine  from 
manganese  dioxide  and  hydrochloric  acid  contain  principally 
manganous  chloride.  Besides  this,  however,  there  is  always  some 
free  acid  and  almost  always  a  considerable  amount  of  ferric  chloride 
present.  The  greater  part  of  the  free  acid  can  be  removed  by 
evaporating  the  solution  until  a  pasty  mass  is  left  which  will 
solidify  on  cooling.  The  iron  can  be  removed  from  the  solution  of 
this  residue  in  virtue  of  the  ease  with  which  ferric  salts  hydrolyze. 
The  nearly  neutral  solution  is  treated  with  suspended  manganous 
carbonate  (obtained  by  treating  a  part  of  the  solution  itself  with 
a  soluble  carbonate).  Ferric  chloride  hydrolyzes  according  to  the 
reversible  reaction,  FeCl3  +  3H20  ^±  3HC1  -f  Fe(OH)3.  In  the 
presence  of  manganous  carbonate  the  small  amount  of  free  acid 
thus  formed  is  continuously  used  up  according  to  the  reaction, 
MnC03  +  2HC1  -»  MnCl2  +  H2O  +  CO2.  Thus  the  reaction  of 
hydrolysis  is  enabled  to  run  to  completion.  The  remaining  solu- 
tion, which  is  almost  absolutely  neutral  and  entirely  free  from  iron 
salts,  yields  crystallized  manganous  chloride,  MnCl2.4H2O,  upon 
evaporation. 

Materials:     waste  liquor  from  chlorine  generator,  500  cc. 

sodium  carbonate. 
Apparatus:  6-inch  evaporating  dish. 

8-inch  evaporating  dish. 

Procedure:  Boil  500  cc.  of  waste  manganese  liquor  in  a  6-inch 
evaporating  dish  under  the  hood  until  the  residue  becomes  pasty. 
After  a  scum  begins  to  form  on  the  surface  of  the  liquid,  there  is 


MANGANOUS  CHLORIDE  295 

danger  of  spattering  and  the  mixture  should  be  stirred  with  a  glass 
rod  until  it  becomes  semi-solid.  Heat  the  residue  to  boiling  with 
1,000  cc.  of  water;  without  filtering  the  solution,  take  one-tenth  of 
it,  dilute  this  portion  to  1,000  cc.,  and  add  a  solution  of  sodium  car- 
bonate to  it  until  all  of  the  manganese  is  precipitated  as  carbonate 
(test  for  complete  precipitation).  Transfer  the  precipitate  to  a 
tall,  common  bottle  and  wash  it  by  decantation  at  least  four 
times.  Add  the  slime  of  manganous  carbonate  to  the  remaining 
nine-tenths  of  the  manganous  chloride  solution,  and  boil  the  mix- 
ture in  a  casserole  until  a  few  drops  of  the  filtered  liquid  give  no  red 
color  when  tested  with  KSCN.  Filter  the  solution  and  evaporate 
it  in  an  8-inch  dish  until  a  crystal  scum  forms  on  blowing  across 
the  surface.  Then  allow  the  solution  to  cool  slowly  and  crystallize, 
leaving  it  for  at  least  12  hours  uncovered  in  a  place  protected  from 
dust.  Collect  the  crystals  and  evaporate  the  mother  liquor  to 
obtain  further  crops  of  crystals  until  practically  all  of  the  salt  has 
crystallized.  Spread  the  light  pink  crystals  on  paper  towels  to  dry, 
and  preserve  the  product  in  an  8-ounce  cork-stoppered  bottle. 

Note.  The  crystals  of  manganous  chloride  are  deliquescent 
when  the  temperature  is  low  and  the  atmosphere  charged  with 
moisture.  If  the  product  cannot  be  obtained  satisfactorily  by  the 
above  directions,  carry  out  the  crystallization  and  drying  in  a  place 
at  a  slightly  elevated  temperature,  25°  to  30° ;  or  cool  the  saturated 
hot  solution  rapidly  by  stirring  or  shaking,  and  dry  the  crystal 
meal  so  obtained  by  rinsing  it  with  alcohol  and  then  letting  the 
latter  evaporate  rapidly. 

QUESTIONS 

1.  Explain  the  purpose  of  the  test  with  potassium  sulpho- 
cyanate. 

2.  Explain  the  action  of  manganese  dioxide  in  the  generation 
of  chlorine  gas  from  hydrochloric  acid.     In  what  state  of  oxidation 
does  manganese  exist  in  the  salt  manganous  chloride? 

3.  If  iron  were  in  the  ferrous  condition,  it  would  not  be  removed 
from  the  solution  by  the  above  procedure.     Explain  why  iron  is 
necessarily  in  the  ferric  condition  in  the  liquors  used. 

4.  Dissolve  a  small  grain  of  manganous  chloride  in  a  half  test 
tube  of  water.     Test  the  solution  with  hydrogen  sulphide;   then 
add  a  few  drops  of  ammonia,  and  if  necessary  pass  in  a  little  more 


296        HEAVY   METALS  OF  GROUPS  VI,   VII,   AND  VIII 

hydrogen  sulphide.  Then  add  acetic  acid  (a  weak  acid)  until  the 
solution  is  again  faintly  acid.  Does  the  manganous  sulphide 
dissolve?  Compare  the  solubility  of  manganous  sulphide  with 
that  of  copper  sulphide;  of  zinc  sulphide. 

5.  Explain  how  facts  involved  in  the  foregoing  preparation 
show  that  Mn(OH)2  is  more  strongly  basic  than  Fe(OH)3. 

PREPARATION  63     - 
POTASSIUM  PERMANGANATE 

Although  manganese  dioxide  is  a  powerful  oxidizing  agent,  it 
is  nevertheless  capable  of  being  itself  oxidized  when  it  is  fused 
with  a  basic  flux.  The  trioxide  of  manganese  is  acidic  in  nature  and 
combines  with  the  base  to  form  a  salt.  Thus  it  is  evident  that  the 
presence  of  a  base  favors  the  oxidation. 

The  dioxide  of  manganese  is  neither  strongly  basic  nor  acidic  in 
nature  and  shows  no  marked  tendency  to  form  salts.  The  mon- 
oxide is  distinctly  basic  and  the  trioxide  is  distinctly  acidic,  so 
that  the  former  forms  salts  with  acids  and  the  latter  with  bases. 
It  follows,  therefore,  that  in  the  presence  of  acids  the  dioxide  has 
a  tendency  to  produce  salts  of  manganous  oxide  whereby  an  atom 
of  oxygen  is  set  free,  and  that  in  the  presence  of  bases  manganese 
dioxide  has  a  tendency  to  take  on  another  atom  of  .oxygen  in  order 
to  produce  a  salt  of  the  trioxide. 

Thus  when  manganese  dioxide  is  fused  with  potassium  hydroxide 
and  an  oxidizing  agent,  the  salt  potassium  manganate  is  formed. 
This  salt  is  soluble  in  water  and  is  fairly  stable  so  long  as  a  con- 
siderable excess  of  potassium  hydroxide  is  present;  but  in  presence 
of  an  acid  —  even  so  weak  a  one  as  carbonic  acid  —  the  manganate 
decomposes  spontaneously,  two-thirds  being  oxidized  to  perman- 
ganate at  the  expense  of  the  other  one-third,  which  is  reduced  again 
to  manganese  dioxide: 

3H2Mn04  -*  2HMnO4  +  MnO2  +  2H2O. 

The  permanganate  (or  permanganic  acid)  corresponds  to  the 
heptoxide  of  manganese,  Mn207,  which  is  the  most  strongly  acid- 
forming  of  the  oxides  of  manganese.  Permanganic  acid  is  a  strong 
and  very  soluble  acid,  it  being  of  approximately  the  same  acid 
strength  as  nitric  or  hydrochloric  acids.  It  is  in  addition  a  very 
powerful  oxidizing  agent. 


POTASSIUM  PERMANGANATE  297 

Materials:     powdered  pyrolusite,  MnO2,  50  grams. 

potassium  hydroxide,  50  grams. 

potassium  chlorate,  25  grams. 

carbon  dioxide. 

shredded  asbestos  for  filter. 
Apparatus:  8  cm.  sheet  iron  crucible. 

furnace  glove. 

iron  stirrer  (old  file  with  wooden  handle). 

tongs. 

suction  filter  and  glass  marble. 

Procedure:  Grind  50  grams  of  pyrolusite  to  as  fine  a  powder  as 
possible  (the  finer  it  is  ground,  the  more  successful  the  preparation). 
Place  50  grams  of  potassium  hydroxide  and  25  grams  of  potassium 
chlorate  in  an  8  cm.  sheet  iron  crucible.  Heat  the  mixture  care- 
fully until  it  is  just  melted.  Remove  the  flame  from  under  the 
crucible  and  add  the  pyrolusite,  a  little  at  a  time,  stirring  vigorously 
all  the  while.  Since  the  charge  in  the  crucible  effervesces  and 
spatters  particles  of  melted  salt,  great  care  should  be  taken  to  keep 
the  eyes  at  a  safe  distance.  The  hand  holding  the  stirrer  should 
be  protected  with  a  thick  glove  and  with  the  other  hand  the 
crucible  should  be  held  firmly  by  means  of  long  iron  tongs.  After 
all  of  the  pyrolusite  is  added,  place  a  small  flame  below  the  crucible, 
and  keep  stirring  the  charge.  Gradually  increase  the  strength  of 
the  flame,  and  stir  continuously  until  the  mass  stiffens  completely. 
Then  cover  the  crucible  and  heat  it  5  minutes  longer  at  a  dull  red 
heat.  When  the  mass  has  cooled,  place  crucible  and  all  in  1  liter 
of  water  in  an  8-inch  porcelain  dish.  After  the  solid  has  entirely 
disintegrated,  remove  the  crucible  and  rinse  it  off  with  a  little 
water  from  the  wash  bottle.  Boil  the  solution  in  the  dish,  and 
at  the  same  time  pass  in  carbon  dioxide  until  the  green  color  of  the 
manganate  has  entirely  changed  to  the  violet-red  color  of  the  per- 
manganate. Test  the  color  by  touching  a  drop  of  the  solution  on 
a  stirring  rod  to  a  piece  of  filter  paper.  If  the  spot  is  violet  with 
no  trace  of  green  and  only  a  fleck  of  brown  manganese  dioxide  in 
the  center,  the  change  to  permanganate  is  complete.  Remove  the 
lamp;  let  the  sludge  settle  in  the  dish  for  5  minutes;  then  pour 
the  solution  through  an  asbestos  filter  (see  Note  4(d)  on  page  8), 
being  careful  to  avoid  stirring  up  the  sludge  until  the  very  last, 
since  the  slimy  precipitate  of  manganese  dioxide  would  so  clog 


298         HEAVY   METALS  OF  GROUPS  VI,  VII,   AND  VIII 

the  filter  as  to  nearly  stop  the  flow.  Lastly,  with  the  aid  of  a  jet 
of  water  from  the  wash  bottle,  transfer  all  the  sludge  to  the  filter 
and  drain  it  free  from  liquid.  Evaporate  the  solution  in  a  clean 
dish  to  a  volume  of  300  cc.  Let  it  settle  a  moment  and  filter  it 
through  asbestos  as  before,  pour  the  filtrate  into  a  6-inch  dish, 
and  allow  it  to  cool  slowly  in  a  place  protected  from  the  dust. 
When  cold,  collect  the  crystals  of  potassium  permanganate  in  a 
filter  funnel  in  which  a  marble  is  placed.  Evaporate  the  mother 
liquor  to  100  cc.,  filter  it  through  asbestos,  and  obtain  a  second 
crop  of  crystals.  Discard  the  remaining  liquid,  since  it  cannot 
contain  more  than  about  6  grams  of  potassium  permanganate  and 
to  evaporate  it  further  would  cause  potassium  chloride  also  to 
crystallize  out.  Weigh  all  the  crystals,  dissolve  them  in  eight 
times  their  weight  of  water  (to  give  a  saturated  solution  at  about 
40°),  filter  the  solution  through  asbestos  at  near  the  boiling  tem- 
perature, and  let  it  cool  slowly  and  crystallize  in  a  small  porcelain 
dish  covered  with  a  watch  glass.  Recover  another  crop  of  crystals 
in  the  same  way  from  the  mother  liquor,  after  evaporating  it  to  a 
volume  of  60  cc.  Allow  the  crystals  to  dry  on  a  clean  unglazed 
plate,  and  preserve  them  in  a  1 -ounce  glass-stoppered  bottle. 

QUESTIONS 

1.  Name  and  give  the  symbols  of  all  the  oxides  of  manganese. 

2.  From  which  oxide  is  K2Mn04  derived?     KMnO4? 

3.  Write  the  reactions  involved  in  the  above  preparation. 

4.  How    could    KMnO4    be    converted    back    into    K2Mn04? 
Equation? 

PREPARATION  64 
MANGANESE  METAL  BY  THE  GOLDSCHMIDT  PROCESS 

The  principle  of  the  production  of  manganese  by  this  process 
is  exactly  the  same  as  that  of  the  production  of  chromium  in 
Prep.  61.  On  account  of  the  violence  of  the  reaction  between  the 
oxide  of  manganese  and  aluminum  it  is  not  advisable  to  ignite  the 
whole  charge  at  once  in  the  crucible;  yet  on  account  of  the  high 
melting  point  of  manganese  a  considerable  quantity  of  charge 
must  be  used  in  order  to  produce  heat  enough  to  obtain  the  metal 
melted  together  in  a  uniform  lump,  instead  of  distributed  in  small 
globules  throughout  the  mass  of  the  slag.  Before  mixing  up  the 


FERROUS  AMMONIUM  SULPHATE  299 

charge,  the  pyrolusite  which  is  used  must  be  first  heated  by  itself 
in  order  to  drive  off  any  water  which  it  may  contain  and  to  convert 
it  to  the  lower  oxide,  Mn3O4. 

Materials:     powdered  pyrolusite,  MnO2,  900  grams. 

granulated  aluminum,  250  grams. 

barium  peroxide,  10  grams. 

magnesium  ribbon,  5  inches. 
Apparatus:  clay  crucible  of  600  cc.  capacity. 

gas  furnace. 

pail  of  dry  sand. 

long-handled  iron  spoon. 

furnace  glove. 

Procedure:  Place  the  pyrolusite  in  a  crucible  and  heat  to  a  bright 
heat  in  a  gas  furnace.  To  prepare  the  charge,  mix  750  grams  of 
this  material,  when  it  is  cooled  sufficiently,  with  250  grams  of 
granulated  aluminum.  Heat  the  empty  crucible  again  in  the 
furnace,  and  while  still  hot  imbed  it  in  a  pail  of  sand.  Place 
about  20  grams  of  the  charge  in  the  bottom  of  the  hot  crucible. 
Put  on  colored  glasses  and  a  heavy  glove;  start  the  reaction  with 
a  fuse  powder  made  of  10  grams  barium  peroxide  and  1  gram  of 
aluminum,  and  a  magnesium  ribbon  (see  Prep.  61),  and  then  add 
fresh  portions  of  the  charge  rapidly  but  without  allowing  the 
reaction  to  become  too  violent.  When  the  crucible  has  cooled, 
break  it,  and  separate  the  regulus  of  metallic  manganese  from  the 
slag  of  fused  aluminum  oxide. 

QUESTIONS 

1.  If  pyrolusite  containing  water  were  used  without  previous 
heating,  what  disadvantage  would  result  during  the  process? 

2.  What  economy  of  materials  is  effected  by  converting  the 
manganese  dioxide  into  the  lower  oxide? 

PREPARATION  65 
FERROUS  AMMONIUM  SULPHATE  FeS04.(NH4)2S04.6H2O 

Corresponding  to  the  two  most  important  oxides  of  iron,  FeO 
and  Fe2O3,  the  two  sulphates,  FeSO4  and  Fe2(SO4)3,  can  be  pre- 
pared. By  dissolving  iron  in  sulphuric  acid  a  solution  of  ferrous 
sulphate  is  obtained.  This,  however,  is  readily  oxidizable,  slowly 
even  by  the  oxygen  of  the  air,  to  the  higher  sulphate,  and  ferrous 


300        HEAVY  METALS  OF  GROUPS  VI,   VII,   AND  VIII 


sulphate  can  only  be  preserved  free  from  ferric  salt  when  all 
oxygen  is  excluded,  or  when  it  is  kept  in  contact  with  an  excess  of 
metallic  iron  in  an  acidified  solution.  Dry  crystallized  ferrous 
sulphate  or  green  vitriol,  FeSO4.7H2O,  can  be  preserved  fairly 
well  without  becoming  oxidized;  but  the  double  ferrous  and  am- 
monium sulphate  is  not  only  more  easily  prepared  on  account  of 
the  readiness  with  which  it  crystallizes,  but  it  is  also  less  easily 
oxidized  by  contact  with  the  air. 

Materials:     crystallized    ferrous    sulphate,    FeSO4.7H20,     70 

grams  =  \  F.W. 
ammonium  sulphate,  (NH4)2SO4,  33  grams. 

Apparatus:  beakers,  filter,  and  crystallizing  dish. 

Procedure:  Prepare  crystallized  ferrous  ammonium  sulphate 
from  ferrous  sulphate  and  ammonium  sulphate.  In  crystallizing 
the  product,  observe  the  solubilities  given  in  the  following  table, 
and  make  use  of  suggestions  given  in  Note  8,  page  13  and  under 
Alum,  page  181: 

A  saturated  solution  contains  for  each  100  grams  of  water  the  given 
number  of  grams  of  the  anhydrous  salt. 


Temperature  

0° 

10° 

20° 

30° 

40° 

50° 

70° 

90° 

FeSO4 

16 

21 

26 

33 

44 

48 

56 

43 

(NH4)2S04  

71 

73 

75 

78 

81 

84 

92 

99 

FeS04.(NH4)2S04  ... 

12 

17 

22 

28 

33 

40 

52 

Test  the  finished  product  for  ferric  iron,  by  dissolving  a  little  in 
water,  adding  a  drop  of  6-n  H2SO4  and  a  few  drops  of  KSCN.  A 
red  color  indicates  ferric  iron. 

QUESTIONS 

1.  What  tests  can  be  applied  to  show  whether  this  preparation 
dissociates  in  solution  into  the  ions  of  the  two  simple  salts? 

2.  Recall  instances  in  which  complex  salts  of  heavy  metals  do 
not  give  the  simple  ions  of  the  heavy  metals. 

PREPARATION  66 
FERRIC  AMMONIUM  ALUM 

In  this  preparation  ferrous  sulphate  is  converted  into  ferric 
sulphate  under  the  oxidizing  action  of  nitric  acid  In  the  presence 
of  the  amount  of  sulphuric  acid  theoretically  necessary  to  form 


FERRIC  AMMONIUM   ALUM  301 

this  salt.  By  the  addition  of  ammonium  sulphate  the  double  salt, 
ferric  ammonium  sulphate,  Fe2(S04)3.(NH4)2SO4.24H2O,  crystal- 
lizes, this  being  one  of  the  isomorphous  series  of  alums  (see  Alum). 

At  25°,  100  grams  of  water  dissolve  44  grams  of  the  anhydrous 
or  124  grams  of  the  hydrated  ferric  ammonium  sulphate. 

Materials:     crystallized    ferrous     sulphate,     FeSO4.7H20,     93 

grams  =  J  F.W. 

ammonium  sulphate,  (NH4)2SO4,  22  grams. 
36-n  H2SO4,  12  cc. 
6-n  HN03, 

Apparatus:  500  cc.  casserole. 

6-inch  evaporating  dish  with  a  watch  glass  or  glass 
plate  for  a  cover. 

Procedure:  Place  the  ferrous  sulphate  and  100  cc.  of  cold  water 
in  the  casserole,  add  the  12  cc.  of  36-n  H2SO4,  warm  to  about  50° 
and  let  the  salt  at  least  partially  dissolve.  Add  20  cc.  6-n  HN03 
and  keep  the  mixture  at  about  50°  until  the  salt  is  entirely  dis- 
solved. Then  heat  the  solution  to  boiling  and  add  6-n  HNO3, 
1  cc.  at  a  time,  until  further  addition  produces  no  further  reaction. 
The  solution  should  now  be  of  a  deep  brownish  yellow  color  but 
perfectly  clear  and  should  give  no  test  for  ferrous  salt.  Test  by 
adding  about  2  drops  to  100  cc.  of  water  and  adding  J  cc.  of 
K3Fe(CN)6.  No  intense  blue  color  should  be  produced,  although 
there  may  be  a  brown  or  yellow  color.  If  a  deep  blue  color  or 
precipitate  does  show  the  presence  of  ferrous  salt  add  more  HNOs 
and  boil  until  the  iron  is  completely  oxidized.  Evaporate  the 
solution  carefully  to  expel  the  greater  part  of  the  excess  of  nitric 
acid.  Keep  the  casserole  in  the  hand,  rotating  the  solution  to  keep 
the  sides  of  the  vessel  wet  with  the  liquid.  It  is  very  important 
not  to  superheat  any  dry  salt  caked  on  the  side  of  the  dish.  When 
the  solution  begins  to  get  thick  and  sticky  stop  heating,  add  water 
to  make  a  volume  of  115  cc.,  heat  to  boiling  and  dissolve  the  am- 
monium sulphate  in  the  hot  solution.  Filter  the  liquid  if  it  is  not 
entirely  clear,  and  set  it  to  crystallize  in  a  6-inch  evaporating  dish, 
covering  the  latter  and  wrapping  it  with  a  towel  so  that  it  may  cool 
slowly.  Large  clear  crystals  should  be  obtained.  Pick  them  out 
by  hand,  rinse  them  quickly  in  a  dish  of  cold  water  and  dry  them 
on  paper  towels.  Obtain  if  possible  a  second  crop  of  crystals  from 
the  mother  liquor.  Preserve  the  product  in  a  10-ounce  cork- 
stoppered  bottle. 


302        HEAVY   METALS  OF  GROUPS   VI,   VII,   AND  VIII 

QUESTIONS 

1.  Write  the  equation  for  the  oxidation  of  ferrous  sulphate  as 
carried  out  in  this  preparation. 

If  an  unacidified  solution  of  ferrous  sulphate  is  oxidized  by  the 
oxygen  of  the  air,  what  products  are  formed?  Equation? 

2.  Write  the  equation  for  the  test  for  ferrous  salt  with  potassium 
ferricyanide. 

3.  Experiment:   Prepare  a  solution  of  a  ferrous  salt  by  dissolv- 
ing 2  grams  of  ferrous  ammonium  sulphate  in  20  cc.  of  water, 
adding  a  little   dilute   sulphuric   acid  and  a  piece  of  iron  wire. 
Test  both  this  solution  and  a  solution  of  a  ferric  salt  (nitrate  or 
chloride)  with  potassium  ferrocyanide,  potassium  ferricyanide,  and 
potassium  sulphocyanate.     Tabulate  the  results.     These  consti- 
tute the  standard  tests  for  ferrous  and  ferric  salts. 

Experiments 

The  elements  of  the  alkali  and  alkaline  earth  families  show  a 
uniform  valence  in  all  their  compounds.  Proceeding  in  the  order 
in  which  the  elements  have  been  taken  up  in  this  book,  a  con- 
stantly increasing  tendency  has  been  shown  to  display  two  or 
more  valences.  In  fact,  the  most  important  chemical  properties 
of  the  elements  considered  in  the  present  chapter  depend  on  their 
ability  to  change  their  valence.  When  the  valence  changes  to  a 
lower  one  the  element  acts  as  an  oxidizing  agent.  Examples  of 
compounds  of  chromium  and  manganese  acting  as  oxidizing  agents 
are  shown  in  Exps.  10  and  23  in  Chap.  IV. 

1.  Stability  of  Carbonates  of  Metals  in  Divalent  State.     Test 
the  stability  of  nickel  carbonate,  NiCO3,  by  heating  1  gram-  of 
it  gently  in  a  test  tube  while  shaking  it  over  a  flame.     Test  the 
gas  evolved  for  carbon  dioxide;    and  compare  the  action  of  the 
remaining  solid,  when  treated  with  hydrochloric  acid,  with  that  of 
the  original  carbonate.     The  carbonates  of  divalent  iron,  cobalt, 
manganese,   and   chromium  are  .all  of  approximately  the  same 
degree  of  stability  as  nickel  carbonate,  so  that  this  one  experi- 
ment may  be  taken  as  typical  of  this  class  of  carbonates. 

2.  To   Show  Whether  the   Carbonate   of  a  Trivalent  Metal 
Can  Exist.     Dissolve  2  grams  of  ferric  alum  in  10  cc.  of  water 
(this  gives  a  trivalent  iron  salt  in  a  solution  that  contains  no 
free  acid).     Add  2-n  Na2C03  slowly  until  no  more  action  takes 


EXPERIMENTS  303 

place.  What  is  the  gas  evolved?  What  is  the  precipitate?  In 
this  experiment  the  ions  Fe+++  and  COs  are  brought  together; 
the  other  ions,  Na+  and  SO4  ,  could  not  react  together  to  give 
any  visible  effect.  If,  therefore,  ferric  carbonate  were  stable  in 
contact  with  water,  it  would  either  form  a  precipitate  if  it  were 
insoluble,  or  if  it  were  soluble  it  would  simply  stay  in  solution  and 
no  effect  would  be  observable.  The  gas  given  off  shows  that  the 
carbonate  is  unstable.  Write  the  equation  for  the  reaction. 
None  of  the  carbonates  of  the  metals  of  this  chapter,  when  they 
are  in  the  trivalent  condition,  are  any  more  stable  than  ferric 
carbonate.  A  salt  of  chromium,  such  as  chromic  alum,  might  be 
used  instead  of  ferric  alum  in  the  above  experiment. 

3.  Oxidation  of  a  Divalent  Oxide.     Heat  \  gram  of  cobalt  car- 
bonate, CoC03,  in  an  open  porcelain  dish,  holding  the  dish  with 
tongs  and  keeping  it  rotating  in  the  flame,  but  not  allowing  the 
porcelain  to  even  approach  a  visible  red  heat.     Heat  until  the 
color    of   the   cobalt   carbonate  has   completely  changed.     Like 
nickel  carbonate,  cobalt  carbonate  is  decomposed  by  heat  into 
cobaltous  oxide,  CoO,  and  carbon  dioxide.     If  the  cobaltous  oxide 
is  readily  oxidized  by  the  oxygen  of  the  air,  it  may  at  once  be 
changed  into  Co2O3  or  Co3O4.     To  test  for  this,  treat  a  little  of  the 
product  with  hydrochloric  acid  in  a  test  tube.     A  clear  solution 
is  obtained,  deep  blue  when  hot  and  concentrated,  pink  when 
cold  and  dilute,  and  chlorine  is  given  off.     The  solution  shows 
the  properties  that  cobaltous  chloride,  CoCl2,  is  known  to  possess. 
Therefore  the  conclusion  is  that  a  higher  oxide  corresponding  to 
CoCl3  (Co3O4  =  CoO.Co203)  was  present  and  that  the  chloride 
formed  by  metathesis  was  unstable  and  decomposed  into  CoCk 
and  free  chlorine. 

Note.  In  Exp.  1  a  higher  oxide  of  nickel  was  probably  formed 
in  the  same  manner,  although  to  a  considerably  less  extent. 

4.  Properties  of  the  Hydroxides.     In  separate  test  tubes,  take 
2  cc.  each  of  normal  solutions  of  (a)  CrCl3,  (b)  MnCl2,  (c)  FeCl2, 
(d)  FeCl3,  (e)  CoCl2,  (/)  NiCl2,  or  of  the  sulphates  or  nitrates  of  the 
same  metals.     Add  10  cc.  of  water  to  each  tube  and  6-n  NaOH 
until  the  solution  is  alkaline  in  each  case. 

(a)  A  light  green  precipitate  (Cr(OH)3)  appears  in  the  first  tube, 
but  it  redissolves  in  an  excess  of  the  reagent,  and  again  precipitates 
when  the  solution  is  heated.  Chromic  hydroxide  is  amphoteric 
and  dissolves  in  NaOH  to  form  sodium  chromite,  Na3CrOa,  or 


304        HEAVY   METALS   OF  GROUPS   VI,   VII,    AND   VIII 

meta-chromite,  NaCrO2,  but  the  acidic  properties  of  Cr(OH)3  are 
so  weak  that  the  chromite  is  completely  hydrolyzed  by  boiling 
water. 

(b)  A  light  buff  colored  precipitate  (Mn(OH)2)  appears  in  the 
second  tube.     It  does- 'not  redissjplve  in  excess  of  reagent,  but,  on 
the  surface  of  the  liquid  in  coniact  with  air,  it  turns  dark  brown. 
Manganous  hydroxide  is  solely  basic;  it  is  easily  oxidized  by  the 
air  to  Mn(OH)3. 

(c)  A  light  greenish  gray  precipitate  appears  in  the  third  tube. 
It  does  not  redissolve  in  excess  of  reagent,  but  at  the  surface  of  the 
liquid  it  turns  reddish  brown.     Ferrous  hydroxide,  Fe(OH)2,  is 
solely  basic;  it  is  oxidized  by  air  to  ferric  hydroxide,  Fe(OH)3. 

(d)  A  reddish  brown  precipitate  (Fe(OH)3)  is  formed  in  the  fourth 
tube.     It  is  unchanged  by  excess  of  reagent. 

(e)  A  precipitate  which  finally  becomes  a  light  brown  tinged 
with  purple,  insoluble  in  excess  of  reagent,  is  formed  in  the  fifth 
tube. 

(/)  An  apple  green  precipitate  insoluble  in  excess  of  reagent  is 
formed  in  the  sixth  tube. 

Save  the  six  tubes  containing  the  precipitates  for  Exp.  5. 

5.  Action  of  Alkaline  Oxidizing  Agents  on  the  Hydroxides. 
Sodium  hypochlorite  is  formed  by  passing  chlorine  into  NaOH. 
Therefore  the  use  of  NaOCl  as  a  reagent  is  equivalent  to  using 
chlorine.  This  reagent  is  made  by  stirring  bleaching  powder  and 
an  excess  of  sodium  carbonate  with  water  and  filtering. 

Shake  each  of  the  tubes  (a)-(/)  from  Exp.  4  to  uniformly 
suspend  the  precipitate  of  hydroxide,  pour  2  cc.  of  each  suspension 
into  a  fresh  test  tube,  and  to  each  add  about  10  cc.  of  the  NaOCl 
reagent.  Observe  the  effect  for  a  minute  and  then  warm  the  tube. 

(a)  The  precipitate  dissolves  and  a  yellow  solution  is  formed. 

2Cr(OH)3  +  3NaOCl  +  4NaOH  ->  2Na2CrO4  +  3NaCl  +  5H2O. 
(6)  The  light  colored  precipitate  turns  dark  brown. 
Mn(OH)2  +  NaOCl ->  Mn02  +  H2O 

Collect  the  dark  brown  precipitate  and  save  it  for  Exp.  6. 
(c)  The  light  greenish  gray  precipitate  turns  reddish  brown. 

2Fe(OH)2  +  NaOCl  +  H2O  ->  2Fe(OH)3  +  NaCl. 


EXPERIMENTS  305 

(d)  The  reddish  brown  Fe(OH)3  is  not  altered. 

(e)  The  light  colored  precipitate  turns  black. 

2Co(OH)2  +  NaOCl  +  H2O  ->  2Co(OH)3  +  NaCl. 

• 

(/)  The  apple  green  precipitate  turns  black. 

2Ni(OH)2  +  NaOCl  +  H2O  -»  Ni(OH)3  +  NaCl. 

6.  Oxidation  in  Alkaline  Fusion,  (a)  Melt  sodium  carbonate 
in  a  loop  on  the  end  of  a  platinum  wire  until  a  colorless  bead  is 
obtained.  Dip  the  bead  in  the  precipitate  of  Mn02  reserved  from 
Exp.  5  (6),  and  melt  it  again  holding  it  in  the  outer  edge  of  the 
flame  to  come  under  the  oxidizing  influence  of  the  air.  (It  is 
better  to  use  a  blow-pipe,  holding  the  bead  in  the  oxidizing  part 
of  the  flame.)  After  the  bead  is  cold  it  has  a  green  color.  The 
oxygen  of  the  air  oxidizes  manganese  to  the  trioxide  in  alkaline 
fusion. 

Mn02  +  i  O2  ->  Mn03 

MnO3  +  NasCOs  ->  Na2Mn04  +  C02. 


(6)  Repeat  (a)  using  Cr(OH)3  instead  of  MnO2.     A  yellow  bead 
is  obtained,  the  chromium  being  oxidized  to  yellow  chromate, 


(c)  Prepare  a  somewhat  larger  amount  of  sodium  manganate  as 
follows:   Melt  a  mixture  of  5  grams  sodium  hydroxide,  1  gram  of 
potassium  nitrate  and  0.1  gram  of  manganese  dioxide  in  a  small 
iron  crucible  and  heat  it  until  it  ceases  to  foam  and  the  crucible 
is  dull  red.     Cool  the  crucible  and  treat  the  contents  with  about 
200  cc.  of  water.     A  most  intense  green  solution  of  sodium  man- 
ganate is  formed.     Let  the  solution  settle  in  a  tall  beaker,  pour 
the  clear,  but  intense  green,  liquid  into  another  beaker,  and  reserve 
it  for  Exp.  7  (a). 

(d)  Repeat  (c)  using  iron  oxide  instead  of  manganese  dioxide. 
The  solution  of  the  melt  is  red,  having  almost  the  same  shade  and 
intensity  of  color  as  permanganate. 

In  strongly  oxidizing  alkaline  fusion,  iron  is  oxidized  to  ferrate 
(Na2FeO4),  which  gives  the  intense  red  color.  This  is  an  unusual 
compound  of  iron,  it  is  formed  only  under  exceptional  conditions, 
and  it  is  very  unstable.  If  any  pink  color  was  observed  in  Exp.  5 
(d)  it  was  due  to  the  formation  of  a  trace  of  ferrate. 

7.  Permanganate,  (a)  Add  6-n  HN03,  drop  by  drop,  to  the 
solution  of  Na2MnO4  reserved  from  Exp.  6  (c).  The  intense  green 


306        HEAVY   METALS   OF  GROUPS  VI,   VII,   AND   VIII 

color  changes  to  the  equally  intense  purplish  red  of  permanganate. 
Let  the  solution  settle  and  carefully  pour  off  the  red  liquid;  there 
is  a  very  small  dark  brown  precipitate  found  in  the  bottom. 
Reserve  the  solution  for  (6). 

Na2MnO4  +  2IINO3  ->  2NaNO3  +  H2Mn04 

3H2MnO4  ->  2HMn04  +  MnO2  I  +  2H2O. 

The  salts  of  manganic  acid,  H2MnO4,  can  exist  in  alkaline  solution, 
but  the  free  manganic  acid  decomposes  at  once.  The  second  of 
the  equations  represents  an  interesting  case  of  oxidation  and  re- 
duction. Manganese  is  the  only  element  whose  valence  changes. 

3MnVI  -*  2Mnvn  +  Mnlv 

the  sum  of  the  valences  of  the  manganese  being  the  same,  namely 
18,  on  each  side  of  the  equation. 

(6)  To  the  red  permanganic  acid  solution  reserved  in  (a),  add 
6-n  NaOH  until  it  is  strongly  alkaline.  The  color  changes  back 
to  the  intense  green  of  the  manganate. 

2MnO4~  +  20H-  ->  2MnO4—  +  H2O  +  JO2. 

The  equation  is  written  ionically  in  order  to  show  that  OH~-ions 
are  used  up  as  the  reaction  progresses,  thus  explaining  why  presence 
of  alkali  favors  the  change  of  permanganate  to  manganate.  Note 
that  the  anions  of  the  two  salts  have  exactly  the  same  composition 
except  for  the  difference  of  one  negative  electron. 

The  oxidation  and  reduction  changes  add  up  as  follows : 

2Mn          +  VII  to  +  VI         2  X  (-  1)  =  -  2 
1  O  -  II  to  0  1  X  (+  2)  =  +  2 

Total  change  =       0 

(c)  To  1  drop  of  Mn(NO3)2  solution  and  10  cc.  6-n  HNO3  in  a 
test  tube  add  1  gram  of  lead  dioxide  and  boil.  Let  the  surplus 
of  lead  dioxide  settle  and  note  that  the  clear  solution  has  the  intense 
permanganate  color. 

Write  the  equation  and  show  that  the  algebraic  sum  of  the 
valence  changes  of  lead  and  manganese  is  zero.  Compare  the 
action  of  lead  dioxide  with  that  of  bismuth  dioxide  in  Chap.  X, 
Exp.  5. 

8.  Chromate  and  Dichromate.  To  a  yellow  solution  of  potas- 
sium chromate  add  6-n  H2S04,  and  note  that  the  color  changes  to 


GENERAL  QUESTIONS  XI  307 

orange  red.  To  an  orange  red  solution  of  potassium  dichromate 
add  6-n  NaOH  and  note  that  the  color  changes  to  yellow. 

The  yellow  color  of  the  chromate  solution  is  due  to  the  Cr04 — 
ion,  the  orange  red  of  the  dichromate  to  the  Cr2O7 —  ion.  The 
changes  take  place  according  to  the  reversible  reaction. 

Cra07—  +  20H-  -*  2CrO4—  +  H2O. 

It  is  obvious  that  presence  of  OH~  ions  must  favor  the  formation 
of  chromate  and  presence  of  H  +  ions,  since  they  remove  OH  -  ions, 
must  favor  the  formation  of  dichromate. 

It  should  be  noted  that  the  valence  of  chromium  is  VI,  both  in 
chromate  and  dichromate.  The  difference  between  manganate 
and  permanganate  lies  in  the  changing  valence  of  the  manganese. 
The  difference  between  chromate  and  dichromate  lies  merely  in 
the  state  of  hydration  and  can  be  referred  back  to  the  acids. 

CrO3  +  H2O  -»  H2CrO4    chromic  acid. 
2Cr03  +  H2O  -»  H2Cr2O7    dichromic  acid. 

GENERAL  QUESTIONS  XI 
HEAVY  METALS  OF  GROUPS  VI,  VII,  AND  VIII 

1.  In  which  groups  of  the  periodic  system  do  the  elements 
chromium,   manganese,   iron,   nickel,   and   cobalt  fall?     What  is 
peculiar  about  the  position  of  the  last  three?     What  other  metals 
belong  to  the  same  family  as  chromium?     In  what  relation  do  they 
stand  to  sulphur,  selenium,  and  tellurium?     In  what  relation  does 
manganese  stand  to  the  halogens?     What  other  elements  occur 
in  the  eighth  group  in  triads  similar  to  iron,  nickel,  and  cobalt? 

2.  How  do  the  monoxides  of  chromium,  manganese,  iron,  cobalt, 
and  nickel  compare  in  basic  strength  with  the  oxides  of  copper 
and  zinc  and  with  the  oxides  of  the  alkali  and  alkaline  earth  metals? 
How  do  the  sesquioxides,  I^Os,  compare  with  the  monoxides  of 
this  group  as  regards  basic  strength? 

What  is  true  as  regards  the  base-  or  acid-forming  properties  of 
the  oxides  higher  than  the  sesquioxides,  e.g.,  of  Cr03,  Mn03, 
Mn207? 

3.  Give  the  formulas  and  names  of  salts  derived  from  each  of 
the  three  oxides  of  chromium,  CrO,  Cr2O3,  CrO3.     In  which  of  its 
compounds   does   chromium  most   resemble   sulphur?   iron   and 
eluminum?  nickel,  cobalt,  copper,  and  zinc? 


308        HEAVY   METALS   OF   GROUPS   VI,  VII,   AND  VIII 

4.  Give  the  formulas  and  names  of  salts  derived  from  each  of 
the  oxides  of  manganese,  MnO,  Mn203,  MnO2,  Mn03,  Mn2O7.     In 
which  of  its  compounds  does  manganese  most  resemble  chlorine? 
aluminum?  cobalt,  nickel,  copper,  and  zinc?  sulphur?  lead  in  the 
dioxide?  /*• 

5.  Give   a  definition   of   oxidation   and   reduction.     Consider 
two  examples,  (a)  the  action  of  KMn04  with  H2SO3,  and  (6)  the 
action  of  K2Cr2Or  with  HC1.     For  both  of  these  cases  write  equa- 
tions according  to  the  oxide  method  as  outlined  in  Question  1,  under 
Prep.  60.     Chromic  Alum.     Also  write  the  equations  according  to 
the  valence  method  as  outlined  in  Question  3  under  the  same  prep- 
aration. 


APPENDIX 
CONCENTRATION  OF  REAGENTS 

The  use  of  a  uniform  concentration  of  6-normal  for  the  acids 
and  bases  in  the  desk  reagent  bottles  has  proved  to  be  very 
satisfactory.  The  more  concentrated  reagents,  which  are  kept 
under  the  hood,  are  of  the  full  strength  supplied  by  the  manu- 
facturers. 

Acetic  acid 

17-normal,  use  commercial  glacial  acetic  acid,  about  99.5%,  of 

sp.  gr.  1.055 

6-normal,  mix  350  cc.  of  glacial  acetic  acid  with  650  cc.  of 
water. 

Hydrochloric  acid 

12-normal,  use  commercial  concentrated  HC1  of  sp.  gr.  1.19 
6-normal,  mix  12-normal  HC1  with  an  equal  volume  of  water 

Nitric  acid 

16-normal,  use  commercial  concentrated  HNO3  of  sp.  gr.  1.42 
6-normal,  mix  380  cc.  of  16-normal  HN03  with  620  cc.  of  water 

Sulphuric  acid 

36-normal,  commercial  96%  H2SO4  of  sp.  gr.  1.84 
6-normal,  pour  1  volume  of  the  96%  H2SO4  into  5  volumes  of 
water 

Ammonium  hydroxide 

15-normal,  use  commercial  concentrated  NH4OH  of  sp.  gr.  0.90 
6-normal,  mix  400  cc.  of  15-normal  NH4OH  with  600  cc.  of 
water 

Sodium  hydroxide 

6-normal,  add  to  250  grams  of  NaOH  enough  water  to  make 
the  volume  1000  cc. 

Salts 

1-normal,  all  salts  solutions  on  the  reagent  shelves  unless  other- 
wise labelled  are  understood  to  be  1-normal 
309 


310  APPENDIX 

TENSION  OF  SATURATED  AQUEOUS  VAPOR 


Temp. 

Pressure 

Temp. 

Pressure 

Temp. 

Pressure 

0° 

4  .  6  mm. 

'  -21° 

v>  18  .  5  mm. 

30° 

31.5  mm. 

5 

6.5 

22 

-    19.7 

35 

41.9 

10 

9.2 

23 

20.9 

40 

55.0 

15 

12.7 

24 

22.2 

50 

92.2 

16 

13.5 

25 

23.6 

60 

149.2 

17 

14.4 

26 

25.1 

'70 

233.8 

18 

15.4 

27 

26.5 

80 

355.5 

19 

16.3 

28 

28.1 

90 

526.0 

20 

17.4 

29 

29.8 

100 

760.0 

ELECTROMOTIVE  SERIES 

When  a  metal  is  in  contact  with  a  solution  containing  its  ions, 
a  difference  in  electrical  potential  arises  due  to  the  resultant  effect 
of  two  tendencies :  —  of  the  metal  on  the  one  hand  to  throw  off  posi- 
tive ions  and  thus  charge  the  solution  positively,  and  of  the  metal 
ions  on  the  other  hand  to  deposit  on  the  metal  and  impart  their 
charges  to  it. 

The  figures  given  in  the  table  are  for  the  potential  of  a  solution, 
normal  in  the  ions  of  the  given  metal,  measured  against  the  metal 
itself  which  dips  in  the  solution.  The  potential  of  a  normal 
solution  of  hydrogen  ions  measured  against  a  hydrogen  electrode 
(platinum  electrode  saturated  with  hydrogen  gas  under  atmos- 
pheric pressure)  is  taken  as  zero  and  the  other  potentials  are 
measured  from  this  arbitrary  zero  point. 

The  elements  for  which  no  figures  are  given  are  placed  in  ap- 
proximately their  correct  position  in  the  series. 

METALS 


K  

,  +2 

0 

Cd 

-j-0 

43 

As  



Na  

+2 

.7 

Fe(Fe++)  

.+0. 

42 

Bi  

— 

Ba.  . 

+2 

7 

Co 

4-0 

9^ 

Sb 



Ca.  . 

+2 

2 

Ni 

4-0 

93 

He;(Hg+) 

-0 

74 

Mg  

+1 

,8 

Sn(Sn++)  

..4-0. 

15 

Ag  

-0. 

80 

Al  

+1, 

3 

Pb. 

.".-. 

..4-0. 

13 

Pd  

— 

Mn.... 

+1 

1 

H?. 

..+0. 

00 

Pt  

— 

Zn 

4-0 

8 

rw 

fln-l-M 

—  n 

32 

Au 

NON-METALS 

F2.... 



T, 

-0 

53 

C12... 

-1.36 

0, 

(in  1-n  OH     solution)  .  . 

-0, 

,40 

O2  (in 

1-n  H+  solution).. 

-1.24 

S. 

— 

Br2.. 

-1.10 

APPENDIX 


311 


1    « 


OK 


S3 


§| 


"    * 


|5 


s 


I- 


S 


sium  2) 
Mg  24.32 


la 


lo 


i: 


Scandium 
Sc  45.1 


ii 

IA 
§ 

Is 

J8 

2x 


is 

B    " 

I 


§!• 


§ 


F 


1s 

22 
I 

Ji 

i 

11 


Columbium 
Cb  93.1 


ll 


n 


Is 

L 
P" 

i 


ll 


IS 


Iff 


e 


II 


I 

IS 


8i 


gi 


14 


312  APPENDIX 

PERIODIC   CLASSIFICATION  OF  THE  ELEMENTS  AC- 
CORDING   TO    THEIR   ATOMIC    NUMBERS    AND 
THE  ARRANGEMENT  OF  THEIR  ELECTRONS 

Atomic  Number.  The  periodic  classification  of  the  elements 
was  originally  made  by  Mendelejeff  on  the  basis  of  atomic  weights. 
But  the  increments  in  atomic  weight  between  successive  elements 
are  not  uniform  and  in  the  cases  of  argon,  cobalt,  and  tellurium 
a  heavier  element  has  to  be  placed  in  the  classification  before  one 
of  smaller  atomic  weight  in  order  to  bring  these  elements  where 
they  obviously  belong  in  virtue  of  their  chemical  properties. 

Another  property  of  the  elements  however  has  been  discovered, 
which  seems  to  be  more  fundamental  even  than  the  atomic  weight 
in  determining  the  progressive  recurrence  of  chemical  properties. 
This  property  has  come  to  be  known  as  atomic  number.  It  is  de- 
termined from  the  characteristic  lines  of  the  X-ray  spectra  of 
the  elements,  and  it  is  found  that  the  reciprocal  of  the  square  root 
of  the  vibration  frequency  of  the  characteristic  X-ray  of  an  element 
always  differs  from  that  of  the  element  of  next  lower  atomic  number 
by  a  constant  quantity.  The  atomic  number  is  therefore  a  strictly 
additive  property;  furthermore,  when  the  elements  are  arranged 
in  the  order  of  their  atomic  numbers,  argon,  cobalt,  and  tellurium 
fall  in  the  places  corresponding  to  their  chemical  properties. 
Giving  hydrogen  the  atomic  number  1,  the  known  elements  satisfy 
all  of  the  atomic  numbers  up  to  and  including  uranium  92,  with  the 
exception  of  43,  61,  75,  85  and  87.  It  is  not  unreasonable  to  sup- 
pose that  elements  having  these  atomic  numbers  exist  and  may 
some  day  be  discovered. 

Theory  of  the  Structure  of  the  Atom.  An  atom  is  believed  to 
be  complex  and  to  consist  of  a  positively  charged  nucleus  sur- 
rounded by  negative  electrons,  the  number  of  which  in  the 
electrically  neutral  atom  is  exactly  equal  to  the  number  of  positive 
charges  of  the  nucleus.  The  -charge  on  the  nucleus  is  an  even 
multiple  of  the  charge  of  a  hydrogen  ion,  and  this  charge,  meas- 
ured in  terms  of  that  of  the  hydrogen  ion  as  a  unit,  is  equal  to 
the  atomic  number.  Confirmation  of  this  view  is  given  by  meas- 
uring the  scattering  effect  on  a  beam  of  alpha  rays  passing 
through  the  elementary  substances. 

According  to  a  theory  proposed  by  Lewis  and  extended  by 
Langmuir,  the  most  stable  configuration  of  electrons  is  possessed 


APPENDIX  313 

by  the  atoms  of  the  inert  gases  helium,  neon,  argon,  krypton, 
xenon  and  nitron.  A  somewhat  less,  but  still  very  stable  con- 
figuration, is  possessed  by  the  beta  forms  of  nickel,  palladium, 
erbium,  and  platinum  (shown  in  the  table  as  Ni/3,  Pd/3,  Er/2, 
Pt/3). 

The  configuration  of  the  atoms  of  the  inert  gases  is  so  stable, 
that  the  atoms  can  neither  part  with  any  of  their  exterior  electrons, 
nor  take  on  added  ones.  These  elements,  do  not  form  any  com- 
pounds and  thus  do  not  exhibit  any  valence. 

The  atoms  of  all  the  other  elements  are  less  stable  and  seek  to 
acquire  the  configuration  of  the  stable  atoms  of  the  inert  gases, 
or  of  the  stable  forms  of  nickel,  palladium,  erbium,  or  platinum. 
This  can  be  accomplished  only  by  losing  some  of  the  exterior 
electrons  or  acquiring  extra  electrons  to  complete  the  outer 
shell  of  the  stable  atoms.  But  this  upsets  the  electrical  neutrality 
of  the  whole  atom  and  makes  it  positive  if  electrons  have  been  lost, 
and  negative,  if  electrons  have  been  acquired.  According  to  the 
theory,  this  tendency  to  change  the  number  of  exterior  electrons 
in  reverting  to  some  one  of  the  stable  forms  of  atoms  is  the  origin 
of  valence  and  the  cause  of  chemical  combination. 

Let  us  choose  for  illustration  the  element  carbon,  atomic  num- 
ber 6.  If  its  atom  loses  four  electrons  it  acquires  the  stable  form 
of  helium,  atomic  number  2,  of  which  the  nuclear  charge  is  2  and 
the  number  of  external  electrons  is  two.  But  the  carbon  atom 
with  a  nuclear  charge  of  6  and  only  two  electrons  will  have  now 
a  surplus  of  four  positive  charges,  and  therefore  a  positive  valence 
of  four,  as  in  carbon  tetrachloride. 

If  the  carbon  atom  takes  on  four  extra  electrons  it  acquires  the 
stable  form  of  neon,  atomic  number  10.  But  the  carbon  atom 
with  a  nuclear  charge  of  only  6  and  ten  external  electrons  has  a 
surplus  of  four  negative  charges,  and  therefore  a  negative  valence 
of  four  as  in  methane,  CTLt. 

The  table  on  the  next  page  shows  the  elements  arranged  in 
periods,  each  of  which  begins  with  an  inert  gas,  and  ends  with  the 
next  inert  gas.  Secondary  periods  within  the  complete  periods  are 
shown,  and  these  arise  from  the  tendency  of  its  atoms  to  revert  to 
the  beta  form  of  the  initial  element  of  the  secondary  period. 

The  elements  which  come  midway  in  the  longer  periods  would 
have  to  lose  or  gain  a  large  number  of  electrons  to  acquire  the 
stable  form  of  the  inert  gases.  It  is  not  surprising  that  this  should 


314 


APPENDIX 


prove  to  be  impossible.  These  elements  show  a  rather  uniform 
tendency  to  develop  the  positive  valence  of  two  or  three.  Their 
atoms  thus  seem  to  be  able  to  throw  off  wo  or  three  electrons 
without  reverting  to  any  one  of  the  completely  stable  forms. 


SH 


ooO 


2* 


%'t 
I*  9 


O 

«c 


3» 


ShH 


35= 


SO 


ni  suoa^oeja  jo  -o^j 


APPENDIX  315 


TABLE  OF  SOLUBILITIES1 

In  the  following  tables  are  given  data  which  should  be  useful  in 
connection  with  the  preparations  and  questions  in  this  book. 

The  formulas  given  are  those  "of  the  crystallized  compounds 
which  most  readily  separate  from  aqueous  solution  at  the  labora- 
tory temperature,  but  it  should  be  remembered  that  many  salts 
have  several  hydrates,  and  it  has  often  been  difficult  to  decide 
which  one  to  place  in  the  table. 

In  the  second  column  the  behavior  of  the  crystallized  salt  when 
it  is  exposed  to  the  air  of  the  laboratory  is  indicated:  s  =  un- 
changed by  exposure  to  atmosphere;  e  =  efflorescent;  d  = 
deliquescent;  d,  e  =  deliquescent  or  efflorescent,  according  as 
to  whether  the  humidity  is  above  or  below  the  average;  CO2  = 
absorbs  carbon  dioxide  and  falls  to  a  white  powder;  Ox  =  com- 
pound is  oxidized,  especially  in  presence  of  moisture. 

In  the  third  column  are  given  the  figures  for  the  solubility  at 
0°,  25°,  and  100°,  except  in  the  cases  in  which  othei  tempera- 
tures are  indicated  in  parenthesis.  Fractions  have,  as  a  rule, 
been  dropped  in  giving  the  solubilities. 

1  Much  of  the  data  in  this  table  has  been  obtained  from  Seidell,  Solu- 
bilities of  Inorganic  and  Organic  Substances. 


316 


APPENDIX 


.                       Formula  of 
crystallized  salt. 

E 

a  -S 

I! 

s  s 
1  ° 

SOLUBILITY  IN  WATER. 

Grams  anhydrous  salt  per 
100  grams  water  in  a  sat- 
urated solution  at 
0°               25°               100° 

F.  W.  per  liter 
of  solution 
at  laboratory 
temperature. 

Aluminum: 
chloride,  A1C13.6H2O  

d 
d 

s 

d 

s 
s 
s 

d 

d 
d 

s 

s 

s 
s 
s 

CO, 

s 

d 

s,  d 

s 
s 

Ox 

d 
d 
d 

s 

e 
d 
e 

s 

(15°)  70 

very  soluble 

31            38          89 

very  soluble 

(io°)  66  (30°)  81  (100°)  128 
12           24 
29            39          77 
118          214        871 

1     hydrolyzes  with  water 
r    to  insoluble  basic  salt; 
J     very  soluble  in  acids 

insoluble;  soluble  in   con- 
centrated acids 

insoluble  in  water  or  acids; 
soluble  in  alkalies 

98            106          149 
0.0023 
32           37         59 
0.0004 

(0°)1.7(25°)4.7(80°)101 

0.008        0.03    0.2 
170          213        272 
5            10          34 
0.00023 
(20°)  0.020   (80°)  0.002 

very  soluble;  hydrolyzes  to 
Ba(SH)2  and  Ba(OH)2 

1  hydrolyzes  with  water  to 
insoluble  basic  salt; 
very  soluble  in  acids 
insoluble  in  acids  or  alkalies 

(00)90(18°)110(li,00)147 

o°)  110  (18°)  127  (60°)  326 
(0°)  76  (40°)  79  (100°)  61 

insoluble;  soluble  in  con- 
centrated acids 

4 
0.8 

6 
2.7 
5 
11 

2.5 
0.00011 
1.7 
0.000015 
0.2 
0.001 
3 
0.3 
0.000010 
0.001 

5 
4.3 
2 

nitrate,  A1(NO3)3.9H2O   .... 
sulphate,  A12(SO4)3.18H2O  .  .  . 

Ammonium: 
acetate,  NH4C2H3O2  
bromide   NH4Br 

(bi)  carbonate,  NH4HCO3.  .  . 
chloride   NH4C1 

nitrate,  NH4NO3    .  . 

Antimony: 
chloride,  SbCls  

sulphate,  Sb2(SO4)8  

sulphide  Sb2Sa 

Arsenic  : 
sulphide,  AsgSa  

Barium: 
bromide,  BaBr2.2H2O  

carbonate,  BaCO3  
chloride,  BaCl2.2H2O  
chromate,  BaCrO4  
hydroxide,  Ba(OH)2.8H2O.  .  . 
iodate,  Ba(IO3)2.H2O  .... 

iodide,  BaI2.6H2O  
nitrate,  Ba(NO3)2  

sulphate,  BaSO4  

sulphite,  BaSO3  

sulphide,  BaS.6H2O 

Bismuth: 
chloride,  Bids  

nitrate,  Bi(NO3)3.5H2O  

sulphate,  Bi2(SO4)3  
sulphide,  Bi2S3  

Cadmium: 
chloride,  CdCl2.2iH2O  . 

nitrate,  Cd(NO3)2.4H2O   . 
sulphate,  CdSO4.2|H2O  

sulphide,  CdS 

APPENDIX 


317 


_  .                     Formula  of 
crystallized  salt. 

Behavior  when 
exposed  to  atmosphere. 

SOLUBILITY  IN  WATER. 

Grams  anhydrous  salt  per 
100  grams  water  in  a  sat- 
urated solution  at 
0°               25°               100° 

F.  W.  per  liter 
of  solution 
at  laboratory 
temperature. 

Calcium: 
acetate,  Ca(C2H3Q2)2.2H2O.  . 
carbonate,  CaCO3 

e 

s 
d 
d 
e 
s 
CO2 
d 
s 
s 

d 

s,  e 

s 
s 
d 

s 

s 

s 
s,d 
e 

•j  melts 
[    38° 

s,  e 
s 

d 

s 
s 
d,  s 

s 

d 

37             34          30 
0.0013 
60          88          159 
(18°)  178 
11            12            3 
0.0016 
0.19        0.16        0.08 
(18°)  122 
0.0007 
0.18        021      016 
0.004 

130 

very  soluble;  melts  36.5° 

120 

insoluble;  soluble  in  acids 

42            53        104 
(o°)84      (91°)  340 
26            39          83 

insoluble  in  water  or  dilute 
acids 

insoluble;  soluble  in  acids 

71           79          108 
82           150        275 

14           23         75 

insoluble  in  water  or  acids 
very  soluble;  melts  35.5° 

0.17          0.35        5.9 
2.0            4.7        27.5 
300 

o°)3.5  (25°)  11.4  (70°)  64 

very  soluble;    melts  37° 

1.9 
0.00013 
5.2 
5.3 
0.8 
0.0002 
0.02 
5.2 
0.00004 
0.015 
0.0003 

8 
4 

3 
4.3 
2 

5.0 

4.8 

1.2 

0.03 
0.6 

0.9 

chloride,  CaCl2.6H2O  
chlorate,  Ca(ClO3)2.2H2O..  .  . 
chromate,  CaCrO4.2H2O  
fluoride,  CaF2  
hydroxide,  Ca(OH)2  
nitrate,  Ca(NO3)2:4H2O  
oxalate,  CaC2O4.H2O  
sulphate,  CaSO4.2H2O  
sulphite,  CaSO3  

Chromium: 
chloride,  CrCl3.6H2O  
nitrate,  Cr(NO3)3.9H2O  
sulphate,  Cr2(SO4)3.18H2O.  .  . 

Cobalt: 
carbonate,  CoCO3  
chloride  CoCl2  6H2O 

nitrate,  Co(NO3)2.6H2O  
sulphate,  CoSO4.7H2O  

sulphide,  CoS  

Copper  : 
carbonate,  CuCOg  

chloride,  CuCl2.2H2O  

nitrate,  Cu(NO3)2.6H2O  
sulphate,  CuSO4.5H2O  

sulphide,  CuS  

Hydrogen: 
arsenic  acid,  H3AsO4.^H2O 
benzoic  acid,  HCrHsC^  
boric  acid,  H3BO3  

iodic  acid,  HIO3  
oxalic  acid,  H2C2O4.2H2O  
phosphoric  acid,  H3PO4  

318 


APPENDIX 


if 

SOLUBILITY  IN  ^ 

VATEB. 

Formula  of 
crystallized  salt. 

Behavior  wl 
expose^  to  atm 

Grams  anhydrous  salt  per 
100  grams  water  in  a  sat- 
urated solution  at 
0°               25°                100° 

F.  W.  per  liter 
of  solution 
at  laboratory 
temperature. 

Iron: 
chloride  (ous),  FeCl2.4H2O.  . 
(ic),  FeCl3.6H2O.  .  .  . 
carbonate  (ous),  FeCO3  
nitrate  (ous),  Fe(NO3)2.6H2O 
(ic),  Fe(NO3)3.9H2O 
sulphate  (ous),  FeSO4.7H2O 
(ic),  Fe2(SO4)3  

d 
d 

s 
Ox 
d 
e,  Ox 
d 

(15°)  67        (80°)  100 
very  soluble;  melts  31° 
insoluble;  soluble  in  acids 
(18°)  82 
very  soluble;  melts  47° 
(0°)  16      (30°)  33      (90°)  43 
very  soluble 

2 

sulphide 

Lead: 
acetate,  Pb(C2H3O2)2.3H2O 
bromide,  PbBr2  

s 

s 
s 

insoluble;  soluble  in  acids 

50        200 
0.5          1.0        4.8 

1.5 
0.02 

carbonate,  PbCO3    

s 

0.0001 

0.000003 

chloride   PbCl2 

s 

0.7          1.1        3.3 

0.05 

hydroxide,  Pb(OH)2  
iodide   PbI2             ... 

s 

0.01 
0.04      0.08        0.44 

0.0004 
0.002 

nitrate,  Pb(NO3)2  

s 

38            59        132 

1.4 

sulphate,  PbSO4  
sulphide,  PbS  

3 

0.004 

insoluble  ;   soluble  in  con- 

0.00013 

Lithium: 
carbonate   Li2CO3 

g 

centrated  strong  acids 

1.5        1.3        0.7 

0.17 

bicarbonate,  LiHCOs  
chloride   LiCl 

S 

d 

(13°)  5.5 
67           82        128 

0.8 
13.3 

hydroxide,  LiOH.H2O  

CO2 

12.7        12.9      17.5 

5.0 

nitrate   LiNO3.3H2O  

d 

(0°)  54  (30°)  138  (70°)  176 

7.3 

sulphate,  Li2SO4  

s 

35            34          30 

2.8 

Magnesium: 
carbonate,  MgCO3.3H2O  
chloride,  MgCl2.6H2O  
hydroxide,  Mg(OH)2  
nitrate,  Mg(NO3)2.6H2O  
sulphate,  MgSO4.7H2O  

Manganese  : 
carbonate,  MnCO3  

s 

d 

C02 
d 
e 

s 

0.2 

53         57           73 
0.001 

(o°)67                   (40°)  85 
27           39         74 

insoluble;  soluble  in  acids 

0.01 
5.1 
0.0002 
4.0 
2.8 

chloride,  MnCl2.4H2O  

e,  d  - 

63           77        115 

5 

nitrate,  Mn(NO3)2.6H2O  

d 

(0°)102  (25°):166  (35.5)331 

5 

sulphate,  MnSO4.5H2O  
sulphide,  MnS  

s,  e 
Ox 

53            65          32 

insoluble;  soluble  in  dilute 

4 

Mercury: 
chloride  (ous),  HgCl  
(ic),  HgCl2  

s 
s 

acids 
0.0002 

3.7          7.4         61 

0.00001 
0.2 

APPENDIX 


319 


„.                        Formula  of 
crystallized  salt. 

Behavior  when 
exposed  to  atmosphere. 

SOLUBILITY  IN  WATER. 

Grams  anhydrous  salt  per 
100  grams  water  in  a  sat- 
urated solution  at 
0°               25°                100° 

F.  W.  per  liter 
of  solution 
at  laboratory 
temperature. 

Mercury: 
iodide  (ic),  Hgl2   .  «. 

S 

Y 

s 

s 
s,d 

s,  d 
e 

s 

d 

s 
s 

d 

s 
s 
s 
s 
s 

d 
d 

s 
s 

d 

s 
s 
s 
s 
s 

d 
d 

s 
s 
s 
s 
s 
s 

0.005 

]  very  soluble  in  a  little  water 
[  much  water  ppts.  basic  salt 
J  very  soluble  in  HNOa 

0.006 

insoluble;  insoluble  in  con- 
centrated acids 

insoluble;  soluble  in  acids 

54            67          88 

(0°)  80  (20°)  96  (95°)  233 
(o°)27   (30°)  43    (99°)  77 

insoluble  in  water  or  dilute 
acids 

5°)  188  (14°)  230  (62°)  492 
3.1          8.0        50 
54            68        104 
89          113        156 
o°)22     (25°)  36    (60°)  60 
3.1         8.2         56 
28            36          57 
59            64          79 
5            16          89 
(18°)  92 
97          119        178 
4.7         9.9         32 
128          148         208 

very  soluble 

13            37          246 

38 
1.9         20 

o°)  2.8  (25°)  8.0  (65°)  25 
7           12         24 
o°)36  (20°)  51  (100°)  122 

very  soluble 
very  soluble 

0°)0.7   (25°)  1.1  (80°)  2.  5 
0.19 
0.00001 
0.003 
15 
0.0002 

0.0001 
0.0001 

4 
6 

2 

25 
0.38 
4.6 
5.9 
2.8 
0.52 
3.9 
2.7 
0.4 
12.4 
18 
0.35 
6.0 

2.6 
1.6 
0.11 
0.33 
0.62 
3.5 

0.06 
0.008 
0.0000006 
0.0001 
0.6 
0.00001 

nitrate  (ous),  HgNO3.H2O.. 
(ic),  Hg(N03)2.iH20 

sulphate  (ous),  Hg2SO4 
sulphide  (ic),  HgS  

Nickel: 
carbonate,  NiCO3  
chloride,  NiCl2.6H2O       . 

nitrate,  NiNO3.6H2O  .    . 

sulphate,  NiSO4.7H2O  .  .  . 

sulphide,  NiS  

Potassium: 
acetate,  KC2H3O2  

brornate,  KBrOa  
bromide,  KBr  
carbonate,  K2CO3.UH2O  
(bi)  carbonate,  KHCO3  
chlorate,  KC1O3  

chloride,  KC1  

chromate,  K2CrO4  

(di)  chromate,  K2Cr2O7  
fluoride,  KF.2H2O  

hydroxide,  KOH.2H2O  
iodate,  KIO3             
iodide,  KI  

manganate,  K2MnO4  
nitrate,  KNO3  
oxalate,  K2C2O4.H2O  

perchlorate,  KC1O4  

permanganate,  KMnO4  

sulphate,  K2SO4  

(bi)  sulphate,  KHSO4  

sulphide,  K2S.5H2O  

sulphite,  K2SO3.2H2O  

Silver: 
acetate,  AgC2H3O2 

bromate,  AgBrOs 

bromide,  AgBr 

carbonate,  Ag2CO3 

chlorate,  AgClOs  

chloride,  AgCl     .           

320 


APPENDIX 


„                           Formula  of 
'                 crystallized  salt*     * 

6 

a! 

if 

|| 

8 

SOLUBILITY  IN  WATEB. 

Grams  anhydrous  salt  per 
100  grams  water  in  a  sat- 
urated solution  at 
0°               25°                100° 

F.  W.  per  liter 
of  solution 
at  laboratory 
temperature. 

Silver: 
chromate,  Ag2CrO4  

s 
d 
s 
s 
s 
s 

d 

s 
s 

s,  e 

s 
s 
e 
s 
s 
s 
e 
d 
s 
d 
e,  d 
s 
s 
d 
e 
d 
fOx 
id,  e 
e 
s,  e 

s 
e 
C02 

e 

s 

f    s 
jOx 

0.002 
(16°)  182 
0.005 
0.0000003 
122          257      952 
0.0025 

very  soluble 
(18°)  0.73     (100°)  1.5 
insoluble  in  water  or  acids 

(0°)34    (25°)  53     (40°)  65 

(5°)  1.3  (30°)  3.9  (100°)  53 
73           87        118 
7.0          28          46 

(0°)6.9      (25°)10     (60°)  16 

82          105        233 
36            36          40 

(0°)32  (21°)  90  (100°)  126 
(0°)  163        (98°)  433 
(21°)  4.  2 
42          114        348 
159          184        302 
73            92        178 
(15°)  3.  2      (100°)  6.  3 

very  soluble 

o°)5.0(32.75°)50.65(ioo°)43 
(25°)  29       (ioo°)  50 

very  soluble 

(0°)  14     (20°)  27     (40°)  50 
(10°)  60  (25°)  76  (45°)  124 

0.001 
44           56        101 
0.4          1.0         32 
40           79        101 
0.01 

(0°)  84         (15°)  270 

0.00015 
13.5 
0.00014 
0.00000001 
8.4 
0.0002 

0.024 

6 

0.15 
6.9 
1.8 
1.1 
6.4 
5.4 
3.3 
5.0 
1.1 
21 
8.1 
7.4 
0.24 

1.2 

2 
5 

0.00007 
3.0 
0.06 
2.7 
0.0006 

7 

fluoride,  AgF  
iodate  AglOs 

iodide  Agl 

nitrate  AgNO3 

oxide,  Ag2O,  dissolves  as  AgOH 
perchiorate  AgClO4 

sulphate,  Ag2SO4  
sulphide,  AgS                     .... 

Sodium: 
acetate,  NaC2H3O2.3H2O  
(tetra)  borate 
(borax),  Na2B4O7.10H2O.,  . 
bromide,  NaBr.2H2O  

carbonate,  Na2CO3.10H2O 
(bi)  carbonate,  NaHCO3  
chlorate,  NaClOs  
chloride   NaCl 

chromate,  Na2CrO4.10H2O.  .  . 
(di)  chromate,Na2Cr2O7.2H2O 
fluoride,  NaF 

hydroxide,  NaOH.H2O   .  . 

iodide,  NaI.2H2O 

nitrate,  NaNO3 

oxalate,  Na2C2O4 

permanganate,  NaMnO4.3H2O 
sulphate,  Na2SO4.10H2O  
(bi)  sulphate,  NaHSO4.H2O.  . 

sulphide,  Na2S.9H2O  

sulphite,  Na2SO3.10H2O  
thiosulphate,  Na2S2O3.5H2O 

Strontium: 
carbonate,  SrCOa    

chloride,  SrCl2.6H2O  

hydroxide,  Sr(OH)2.8H2O.... 
nitrate   Sr(NO3)2  4H2O 

sulphate  SrSO4 

Tin: 
chloride  (ous)  SnCl2.2H2O.  .  . 

APPENDIX 


321 


Formula  of 
crystallized  salt. 

Behavior  when 
exposed  to  atmosphere. 

SOLUBILITY  IN  WATER. 

Grams  anhydrous  salt  per 
100  grams  water  in  a  sat- 
urated solution  at 
0°               25°               100° 

F.  W.  per  liter 
of  solution 
at  laboratory 
temperature. 

Zinc: 
carbonate,  ZnCOs  

S 
d 
d 

e 

s 

0.004? 
208          432        615 
95          127 
42            58          81 

insoluble;  soluble  in  acids 

0.0003? 
9.2 
4.7 
3.1 

chloride,  ZnCl2.3H2O  

nitrate,  Zn(NO3)2.6H2O  

sulphate,  ZnSO4.7H2O  

sulphide,  ZnS  

7E  ON  THE 


